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C Apitiude Questions and Ans RRS feed

  • Question

  • I think this thread helps you a lot ..

     

    TC

    Thursday, May 24, 2007 1:25 PM

Answers

  • 1.      void main()

    {

                int  const * p=5;

                printf("%d",++(*p));

    }

    Answer:

                            Compiler error: Cannot modify a constant value.

    Explanation:   

    p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

    Thursday, May 24, 2007 1:29 PM
  • 1.      main()

    {

                char s[ ]="man";

                int i;

                for(i=0;s[ i ];i++)

                printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),iSleep);

    }

    Answer:

                            mmmm

                           aaaa

                           nnnn

    Explanation:

    sIdea, *(i+s), *(s+i), iSleep are all different ways of expressing the same idea. Generally  array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as sIdea. iSleep may be surprising. But in the  case of  C  it is same as sIdea.

    Thursday, May 24, 2007 1:32 PM
  • 1.      main()

    {

                float me = 1.1;

                double you = 1.1;

                if(me==you)

    printf("I love U");

    else

                            printf("I hate U");

    }

    Answer:

    I hate U

    Explanation:

    For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value  represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.

     

    Thursday, May 24, 2007 1:34 PM
  • 1.      main()

                {

                static int var = 5;

                printf("%d ",var--);

                if(var)

                            main();

                }

    Answer:

    5 4 3 2 1

                Explanation:

    When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively. 

    Thursday, May 24, 2007 1:35 PM
  • 5.      main()

    {

                 int c[ ]={2.8,3.4,4,6.7,5};

                 int j,*p=c,*q=c;

                 for(j=0;j<5;j++) {

                            printf(" %d ",*c);

                            ++q;     }

                 for(j=0;j<5;j++){

    printf(" %d ",*p);

    ++p;     }

    }

     

    Answer:

                            2 2 2 2 2 2 3 4 6 5

                Explanation:

    Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

    Thursday, May 24, 2007 1:36 PM
  • 6.      main()

    {

                extern int i;

                i=20;

    printf("%d",i);

    }

     

    Answer: 

    Linker Error : Undefined symbol '_i'

    Explanation:

                            extern storage class in the following declaration,

                                        extern int i;

    specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

     

    Thursday, May 24, 2007 1:37 PM
  • 7.      main()

    {

                int i=-1,j=-1,k=0,l=2,m;

                m=i++&&j++&&k++||l++;

                printf("%d %d %d %d %d",i,j,k,l,m);

    }

    Answer:

                            0 0 1 3 1

    Explanation :

    Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression  i++ && j++ && k++’ is executed first. The result of this expression is 0    (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

    Thursday, May 24, 2007 1:38 PM
  • 8.      main()

    {

                char *p;

                printf("%d %d ",sizeof(*p),sizeof(p));

    }

     

    Answer:

                            1 2

    Explanation:

    The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

    Thursday, May 24, 2007 1:39 PM
  • 9.      main()

    {

                int i=3;

                switch(i)

                 {

                    defaultStick out tonguerintf("zero");

                    case 1: printf("one");

                               break;

                   case 2Stick out tonguerintf("two");

                              break;

                  case 3: printf("three");

                              break;

                  } 

    }

    Answer :

    three

    Explanation :

    The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

    Thursday, May 24, 2007 1:40 PM
  • cool, start at beginners level questions and then end with expert level questions, it would be good to have t...
    Thursday, May 24, 2007 1:40 PM
  • 10.      main()

    {

                  printf("%x",-1<<4);

    }

    Answer:

    fff0

    Explanation :

    -1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

    Thursday, May 24, 2007 1:41 PM
  • 11.      main()

    {

                  printf("%x",-1<<4);

    }

    Answer:

    fff0

    Explanation :

    -1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

    Thursday, May 24, 2007 1:42 PM
  • 11.      main()

    {

                char string[]="Hello World";

                display(string);

    }

    void display(char *string)

    {

                printf("%s",string);

    }

                Answer:

    Compiler Error : Type mismatch in redeclaration of function display

                Explanation :

    In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and

    return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

    Thursday, May 24, 2007 3:30 PM
  • 12.      main()

    {

                int c=- -2;

                printf("c=%d",c);

    }

    Answer:

                                        c=2;

                Explanation:

    Here unary minus (or negation) operator is used twice. Same maths  rules applies, ie. minus * minus= plus.

    Note:

    However you cannot give like --2. Because -- operator can  only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

    Thursday, May 24, 2007 3:30 PM
  • 13.      #define int char

    main()

    {

                int i=65;

                printf("sizeof(i)=%d",sizeof(i));

    }

    Answer:

                            sizeof(i)=1

    Explanation:

    Since the #define replaces the string  int by the macro char

    Thursday, May 24, 2007 3:31 PM
  • 14.      main()

    {

    int i=10;

    i=!i>14;

    Printf ("i=%d",i);

    }

    Answer:

    i=0

     

     

                Explanation:

    In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol.  ! is a unary logical operator. !i (!10) is 0 (not of true is false).  0>14 is false (zero).

    Thursday, May 24, 2007 3:32 PM
  • 15.      #include<stdio.h>

    main()

    {

    char s[]={'a','b','c','\n','c','\0'};

    char *p,*str,*str1;

    p=&s[3];

    str=p;

    str1=s;

    printf("%d",++*p + ++*str1-32);

    }

    Answer:

    77       

    Explanation:

    p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.

     Now performing (11 + 98 – 32), we get 77("M");

     So we get the output 77 :: "M" (Ascii is 77).

    Thursday, May 24, 2007 3:32 PM
  • 16.      #include<stdio.h>

    main()

    {

    int a[2][2][2] = { {10,2,3,4}, {5,6,7,8}  };

    int *p,*q;

    p=&a[2][2][2];

    *q=***a;

    printf("%d----%d",*p,*q);

    }

    Answer:

    SomeGarbageValue---1

    Explanation:

    p=&a[2][2][2]  you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

               

    Thursday, May 24, 2007 3:33 PM
  • 17.      #include<stdio.h>

    main()

    {

    struct xx

    {

          int x=3;

          char name[]="hello";

     };

    struct xx *s;

    printf("%d",s->x);

    printf("%s",s->name);

    }

                Answer:

    Compiler Error

    Explanation:

    You should not initialize variables in declaration

     

    Thursday, May 24, 2007 3:33 PM
  • 18.      #include<stdio.h>

    main()

    {

    struct xx

    {

    int x;

    struct yy

    {

    char s;

                struct xx *p;

    };

    struct yy *q;

    };

    }

    Answer:

    Compiler Error

    Explanation:

    The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

    Thursday, May 24, 2007 3:34 PM
  • 19.      main()

    {

    printf("\nab");

    printf("\bsi");

    printf("\rha");

    }

    Answer:

    hai

    Explanation:

    \n  - newline

    \b  - backspace

    \r  - linefeed

    Thursday, May 24, 2007 3:34 PM
  • 20.      main()

    {

    int i=5;

    printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);

    }

    Answer:

    45545

    Explanation:

    The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the  evaluation is from right to left, hence the result.

    Thursday, May 24, 2007 3:35 PM
  • 21.      #define square(x) x*x

    main()

    {

    int i;

    i = 64/square(4);

    printf("%d",i);

    }

    Answer:

    64

    Explanation:

    the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

     

    Thursday, May 24, 2007 3:35 PM
  • 22.      main()

    {

    char *p="hai friends",*p1;

    p1=p;

    while(*p!='\0') ++*p++;

    printf("%s   %s",p,p1);

    }

    Answer:

    ibj!gsjfoet

                Explanation:

                            ++*p++ will be parse in the given order

    Ø      *p that is value at the location currently pointed by p will be taken

    Ø      ++*p the retrieved value will be incremented

    Ø      when ; is encountered the location will be incremented that is p++ will be executed

    Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print anything
    Thursday, May 24, 2007 3:36 PM
  • 23.      #include <stdio.h>

    #define a 10

    main()

    {

    #define a 50

    printf("%d",a);

    }

    Answer:

    50

    Explanation:

    The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

    Thursday, May 24, 2007 3:37 PM
  • 1.      #define clrscr() 100

    main()

    {

    clrscr();

    printf("%d\n",clrscr());

    }

    Answer:

    100

    Explanation:

    Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input  program to compiler looks like this :

                            main()

                            {

                                 100;

                                 printf("%d\n",100);

                            }

                Note:  

    100; is an executable statement but with no action. So it doesn't give any problem

    Thursday, May 24, 2007 3:38 PM
  • 25.      main()

    {

    printf("%p",main);

    }

    Answer:

                            Some address will be printed.

    Explanation:

                Function names are just addresses (just like array names are addresses).

    main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

     

    Thursday, May 24, 2007 3:38 PM
  • 27

                main()

    {

    clrscr();

    }

    clrscr();

               

    Answer:

    No output/error

    Explanation:

    The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).

    Thursday, May 24, 2007 3:39 PM
  • 28 

               enum colors {BLACK,BLUE,GREEN}

     main()

    {

     

     printf("%d..%d..%d",BLACK,BLUE,GREEN);

      

     return(1);

    }

    Answer:

    0..1..2

    Explanation:

    enum assigns numbers starting from 0, if not explicitly defined.

    Thursday, May 24, 2007 3:40 PM
  • 29

    void main()

    {

     char far *farther,*farthest;

     

     printf("%d..%d",sizeof(farther),sizeof(farthest));

      

     }

    Answer:

    4..2 

    Explanation:

                the second pointer is of char type and not a far pointer

    Thursday, May 24, 2007 3:40 PM
  • 30

    main()

    {

     int i=400,j=300;

     printf("%d..%d");

    }

    Answer:

    400..300

    Explanation:

    printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program,then printf will take garbage values.

     

    Thursday, May 24, 2007 3:41 PM
  • 31

     main()

    {

     char *p;

     p="Hello";

     printf("%c\n",*&*p);

    }

    Answer:

    H

    Explanation:

    * is a dereference operator & is a reference  operator. They can be    applied any number of times provided it is meaningful. Here  p points to  the first character in the string "Hello". *p dereferences it and so its value is H. Again  & references it to an address and * dereferences it to the value H.

    Thursday, May 24, 2007 3:42 PM
  • 32

    main()

    {

        int i=1;

        while (i<=5)

        {

           printf("%d",i);

           if (i>2)

                  goto here;

           i++;

        }

    }

    fun()

    {

       here:

         printf("PP");

    }

    Answer:

    Compiler error: Undefined label 'here' in function main

    Explanation:

    Labels have functions scope, in other words The scope of the labels is limited to functions . The label 'here' is available in function fun() Hence it is not visible in function main.

    Thursday, May 24, 2007 3:42 PM
  • 33

    main()

    {

       static char names[5][20]={"pascal","ada","cobol","fortran","perl"};

        int i;

        char *t;

        t=names[3];

        names[3]=names[4];

        names[4]=t; 

        for (i=0;i<=4;i++)

                printf("%s",namesIdea);

    }

    Answer:

    Compiler error: Lvalue required in function main

    Explanation:

    Array names are pointer constants. So it cannot be modified.

    Thursday, May 24, 2007 3:43 PM
  • 34

    void main()

    {

                int i=5;

                printf("%d",i++ + ++i);

    }

    Answer:

    Output Cannot be predicted  exactly.

    Explanation:

    Side effects are involved in the evaluation of   i

    Thursday, May 24, 2007 3:44 PM
  • 35

    void main()

    {

                int i=5;

                printf("%d",i+++++i);

    }

    Answer:

    Compiler Error

    Explanation:

    The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.

    Thursday, May 24, 2007 3:45 PM
  • 36

    #include<stdio.h>

    main()

    {

    int i=1,j=2;

    switch(i)

     {

     case 1:  printf("GOOD");

                    break;

     case j:  printf("BAD");

                   break;

     }

    }

    Answer:

    Compiler Error: Constant expression required in function main.

    Explanation:

    The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).

                Note:

    Enumerated types can be used in case statements.

     

    Thursday, May 24, 2007 3:48 PM
  • 37

     

    main()

    {

    int i;

    printf("%d",scanf("%d",&i));  // value 10 is given as input here

    }

    Answer:

    1

    Explanation:

    Scanf returns number of items successfully read and not 1/0.  Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

    Thursday, May 24, 2007 3:49 PM
  • 38

    #define f(g,g2) g##g2

    main()

    {

    int var12=100;

    printf("%d",f(var,12));

                }

    Answer:

    100

    Thursday, May 24, 2007 3:49 PM
  • 39

    main()

    {

    int i=0;

     

    for(;i++;printf("%d",i)) ;

    printf("%d",i);

    }

    Answer:

                1

    Explanation:

    before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

     

    Thursday, May 24, 2007 3:50 PM
  • 40

    #include<stdio.h>

    main()

    {

      char s[]={'a','b','c','\n','c','\0'};

      char *p,*str,*str1;

      p=&s[3];

      str=p;

      str1=s;

      printf("%d",++*p + ++*str1-32);

    }

    Answer:

    M

    Explanation:

    p is pointing to character '\n'.str1 is pointing to character 'a' ++*p meAnswer:"p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10. then it is incremented to 11. the value of ++*p is 11. ++*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is subtracted from 32.

    i.e. (11+98-32)=77("M");

    Thursday, May 24, 2007 3:51 PM
  • 41

    #include<stdio.h>

    main()

    {

      struct xx

       {

          int x=3;

          char name[]="hello";

       };

    struct xx *s=malloc(sizeof(struct xx));

    printf("%d",s->x);

    printf("%s",s->name);

    }

    Answer:

    Compiler Error

    Explanation:

    Initialization should not be done for structure members inside the structure declaration

    Thursday, May 24, 2007 3:52 PM
  • 42

    #include<stdio.h>

    main()

    {

    struct xx

     {

                  int x;

                  struct yy

                   {

                     char s;

                     struct xx *p;

                   };

                             struct yy *q;

                };

                }

    Answer:

    Compiler Error

    Explanation:

    in the end of nested structure yy a member have to be declared.

    Thursday, May 24, 2007 3:53 PM
  • 44

    main()

    {

    printf("%d", out);

    }

    int out=100;

    Answer:

    Compiler error: undefined symbol out in function main.

    Explanation:

    The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

    Thursday, May 24, 2007 3:55 PM
  • 45

    main()

    {

     extern out;

     printf("%d", out);

    }

     int out=100;

    Answer:

    100     

                Explanation: 

    This is the correct way of writing the previous program.

    Thursday, May 24, 2007 3:55 PM
  • 46

    main()

    {

     show();

    }

    void show()

    {

     printf("I'm the greatest");

    }

    Answer:

    Compier error: Type mismatch in redeclaration of show.

    Explanation:

    When the compiler sees the function show it doesn't know anything about it. So the default return type (ie, int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error.

    The solutions are as follows:

    1. declare void show() in main() .

    2. define show() before main().

    3. declare extern void show() before the use of show().

    Thursday, May 24, 2007 3:56 PM
  • 47

     

    main( )

    {

      int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};

      printf(“%u %u %u %d \n”,a,*a,**a,***a);

            printf(“%u %u %u %d \n”,a+1,*a+1,**a+1,***a+1);

           }

    Answer:

    100, 100, 100, 2

    114, 104, 102, 3

    Explanation:

                      The given array is a 3-D one. It can also be viewed as a 1-D array.

                                                                                                                                                                                                                                                                         

    2

    4

    7

    8

    3

    4

    2

    2

    2

    3

    3

    4

       100  102  104  106 108   110  112  114  116   118   120   122

     

    thus, for the first printf statement a, *a, **a  give address of  first element . since the indirection ***a gives the value. Hence, the first line of the output.

    for the second printf a+1 increases in the third dimension thus points to value at 114, *a+1 increments in second dimension thus points to 104, **a +1 increments the first dimension thus points to 102 and ***a+1 first gets the value at first location and then increments it by 1. Hence, the output.

    Thursday, May 24, 2007 3:57 PM
  • 48

    main( )

    {

      int a[ ] = {10,20,30,40,50},j,*p;

      for(j=0; j<5; j++)

        {

    printf(“%d” ,*a);

    a++;

        }

        p = a;

       for(j=0; j<5; j++)

          {

    printf(“%d ” ,*p);

    p++;

          }

     }

    Answer:

    Compiler error: lvalue required.

                           

    Explanation:

    Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

    Thursday, May 24, 2007 3:57 PM
  • 49

    main( )

    {

     static int  a[ ]   = {0,1,2,3,4};

     int  *p[ ] = {a,a+1,a+2,a+3,a+4};

     int  **ptr =  p;

     ptr++;

     printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);

     *ptr++;

     printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);

     *++ptr;

     printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);

     ++*ptr;

           printf(“\n %d  %d  %d”, ptr-p, *ptr-a, **ptr);

    }

    Answer:

                111

                222

                333

                344

    Explanation:

    Let us consider the array and the two pointers with some address

    a   

    0

    1

    2

    3

    4

       100      102      104      106      108

                                                               p

    100

    102

    104

    106

    108

                                           1000    1002    1004    1006    1008

               ptr 

    1000

    2000

    After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1,  *ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102  so the value is (102 – 100)/(scaling factor) = 1,  **ptr is the value stored in the location pointed by  the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is  1, 1, 1.

    After execution of *ptr++ increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the second printf are ptr – p = 2, *ptr – a = 2, **ptr = 2.

    After execution of *++ptr increments value of the value in ptr by scaling factor, so it becomes1004. Hence, the outputs for the third printf are ptr – p = 3, *ptr – a = 3, **ptr = 3.

    After execution of ++*ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 10 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.

    Thursday, May 24, 2007 3:58 PM
  • 50

    main( )

    {

     char  *q;

     int  j;

     for (j=0; j<3; j++) scanf(“%s” ,(q+j));

     for (j=0; j<3; j++) printf(“%c” ,*(q+j));

     for (j=0; j<3; j++) printf(“%s” ,(q+j));

    }

    Explanation:

    Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1. Suppose the inputs are MOUSE,  TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as

    M

    O

    U

    S

    E

    \0

    When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.

    M

    T

    R

    A

    C

    K

    \0

    The third input  starts filling from the location 102

    M

    T

    V

    I

    R

    T

    U

    A

    L

    \0

    This is the final value stored .

    The first printf prints the values at the position q, q+1 and q+2  = M T V

    The second printf prints three strings starting from locations q, q+1, q+2

     i.e  MTVIRTUAL, TVIRTUAL and VIRTUAL.

    Thursday, May 24, 2007 3:58 PM
  • 51

    main( )

    {

     void *vp;

     char ch = ‘g’, *cp = “goofy”;

     int j = 20;

     vp = &ch;

     printf(“%c”, *(char *)vp);

     vp = &j;

     printf(“%d”,*(int *)vp);

     vp = cp;

     printf(“%s”,(char *)vp + 3);

    }

    Answer:

                g20fy

    Explanation:

    Since a void pointer is used it can be type casted to any  other type pointer. vp = &ch  stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly  the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

    Thursday, May 24, 2007 5:27 PM
  • 52

    main ( )

    {

     static char *s[ ]  = {“black”, “white”, “yellow”, “violet”};

     char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;

     p = ptr;

     **++p;

     printf(“%s”,*--*++p + 3);

    }

    Answer:

                ck

    Explanation:

    In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1 , thus now value of p =  s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

    Thursday, May 24, 2007 5:28 PM
  • 53

    main()

    {

     int  i, n;

     char *x = “girl”;

     n = strlen(x);

     *x = xNo;

     for(i=0; i<n; ++i)

       {

    printf(“%s\n”,x);

    x++;

       }

     }

    Answer:

    (blank space)

    irl

    rl

    l

     

    Explanation:

    Here a string (a pointer to char) is initialized with a value “girl”.  The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘\0’) to the first location. Now the string becomes “\0irl” . Now the printf statement prints the string after each iteration it increments it starting position.  Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e “irl” and the third time it prints “rl” and the last time it prints “l” and the loop terminates.

    Thursday, May 24, 2007 5:28 PM
  • 54

    int i,j;

                for(i=0;i<=10;i++)

                {

                j+=5;

                assert(i<5);

                }

    Answer:

    Runtime error: Abnormal program termination.

                                        assert failed (i<5), <file name>,<line number>

    Explanation:

    asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use,

                #undef NDEBUG

    and this will disable all the assertions from the source code. Assertion

    is a good debugging tool to make use of. 

     

    Thursday, May 24, 2007 5:29 PM
  • 55

    main()

                {

                int i=-1;

                +i;

                printf("i = %d, +i = %d \n",i,+i);

                }

    Answer:

     i = -1, +i = -1

    Explanation:

    Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

    Thursday, May 24, 2007 5:29 PM
  • 56

     

    What are the files which are automatically opened when a C file is executed?

    Answer:

    stdin, stdout, stderr (standard input,standard output,standard error).

    Thursday, May 24, 2007 5:30 PM
  • 57 what will be the position of the file marker?

                a: fseek(ptr,0,SEEK_SET);

                b: fseek(ptr,0,SEEK_CUR);

     

    Answer :

                a: The SEEK_SET sets the file position marker to the starting of the file.

                            b: The SEEK_CUR sets the file position marker to the current position

                of the file.

    Thursday, May 24, 2007 5:31 PM
  • 58

    main()

                {

                char name[10],s[12];

                scanf(" \"%[^\"]\"",s);

                }

                How scanf will execute?

    Answer:

    First it checks for the leading white space and discards it.Then it matches with a quotation mark and then it  reads all character upto another quotation mark.

    Thursday, May 24, 2007 5:31 PM
  • 59

    What is the problem with the following code segment?

                while ((fgets(receiving array,50,file_ptr)) != EOF)

                                        ;

    Answer & Explanation:

    fgets returns a pointer. So the correct end of file check is checking for != NULL.

     

    Thursday, May 24, 2007 5:32 PM
  • 60

    main()

                {

                main();

                }

    Answer:

     Runtime error : Stack overflow.

    Explanation:

    main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.

     

    Thursday, May 24, 2007 5:32 PM
  • 61

    main()

                {

                char *cptr,c;

                void *vptr,v;

                c=10;  v=0;

                cptr=&c; vptr=&v;

                printf("%c%v",c,v);

                }

    Answer:

    Compiler error (at line number 4): size of v is Unknown.

    Explanation:

    You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

    Thursday, May 24, 2007 5:33 PM
  • 62

    main()

                {

                char *str1="abcd";

                char str2[]="abcd";

                printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));

                }

    Answer:

    2 5 5

    Explanation:

    In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.

    Thursday, May 24, 2007 5:33 PM
  •  

    63 main()

                {

                char not;

                not=!2;

                printf("%d",not);

                }

    Answer:

    0

    Explanation:

    ! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

    Thursday, May 24, 2007 5:34 PM
  • 64

    #define FALSE -1

                #define TRUE   1

                #define NULL   0

                main() {

                   if(NULL)

                            puts("NULL");

                   else if(FALSE)

                            puts("TRUE");

                   else

                            puts("FALSE");

                   }

    Answer:

    TRUE

    Explanation:

    The input program to the compiler after processing by the preprocessor is,

                main(){

                            if(0)

                                        puts("NULL");

                else if(-1)

                                        puts("TRUE");

                else

                                        puts("FALSE");

                            }

    Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.

    Thursday, May 24, 2007 5:35 PM
  • 65

    main()

                {

                int k=1;

                printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");

                }

    Answer:

    1==1 is TRUE

    Explanation:

    When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".

    Thursday, May 24, 2007 5:36 PM
  • 66

    main()

                {

                int y;

                scanf("%d",&y); // input given is 2000

                if( (y%4==0 && y%100 != 0) || y%100 == 0 )

                     printf("%d is a leap year");

                else

                     printf("%d is not a leap year");

                }

    Answer:

    2000 is a leap year

    Explanation:

    An ordinary program to check if leap year or not.

     

    Thursday, May 24, 2007 5:36 PM
  • 67

    #define max 5

                #define int arr1[max]

                main()

                {

                typedef char arr2[max];

                arr1 list={0,1,2,3,4};

                arr2 name="name";

                printf("%d %s",list[0],name);

                }

    Answer:

    Compiler error (in the line arr1 list = {0,1,2,3,4})

    Explanation:

    arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.

    Rule of Thumb:

    #defines are used for textual replacement whereas typedefs are used for declaring new types.

    Thursday, May 24, 2007 5:37 PM
  • 68

    int i=10;

                main()

                {

                 extern int i;

                  {

                     int i=20;

                            {

                             const volatile unsigned i=30;

                             printf("%d",i);

                            }

                      printf("%d",i);

                   }

                printf("%d",i);

                }

    Answer:

    30,20,10

    Explanation:

    '{' introduces new block and thus new scope. In the innermost block i is declared as,

                const volatile unsigned

    which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.

     

    Thursday, May 24, 2007 5:37 PM
  • 69

    main()

                {

                    int *j;

                    {

                     int i=10;

                     j=&i;

                     }

                     printf("%d",*j);

    }

    Answer:

    10

    Explanation:

    The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

    Thursday, May 24, 2007 5:38 PM
  • 70

    main()

                {

                int i=-1;

                -i;

                printf("i = %d, -i = %d \n",i,-i);

                }

    Answer:

    i = -1, -i = 1

    Explanation:

    -i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

    Thursday, May 24, 2007 5:38 PM
  • 71

    #include<stdio.h>

    main()

     {

       const int i=4;

       float j;

       j = ++i;

       printf("%d  %f", i,++j);

     }

    Answer:

    Compiler error

                Explanation:

    i is a constant. you cannot change the value of constant

    Thursday, May 24, 2007 5:39 PM
  • 72

    #include<stdio.h>

    main()

    {

      int a[2][2][2] = { {10,2,3,4}, {5,6,7,8}  };

      int *p,*q;

      p=&a[2][2][2];

      *q=***a;

      printf("%d..%d",*p,*q);

    }

    Answer:

    garbagevalue..1

    Explanation:

    p=&a[2][2][2]  you declare only two 2D arrays. but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. now q is pointing to starting address of a.if you print *q meAnswer:it will print first element of 3D array.

    Thursday, May 24, 2007 5:40 PM
  • 73

    #include<stdio.h>

    main()

      {

        register i=5;

        char j[]= "hello";                    

         printf("%s  %d",j,i);

    }

    Answer:

    hello 5

    Explanation:

    if you declare i as register  compiler will treat it as ordinary integer and it will take integer value. i value may be  stored  either in register  or in memory.

     

    Thursday, May 24, 2007 5:40 PM
  • 74

    main()

    {

                  int i=5,j=6,z;

                  printf("%d",i+++j);

                 }

    Answer:

    11

    Explanation:

    the expression i+++j is treated as (i++ + j)   

    Thursday, May 24, 2007 5:41 PM
  • 76

    struct aaa{

    struct aaa *prev;

    int i;

    struct aaa *next;

    };

    main()

    {

     struct aaa abc,def,ghi,jkl;

     int x=100;

     abc.i=0;abc.prev=&jkl;

     abc.next=&def;

     def.i=1;def.prev=&abc;def.next=&ghi;

     ghi.i=2;ghi.prev=&def;

     ghi.next=&jkl;

     jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;

     x=abc.next->next->prev->next->i;

     printf("%d",x);

    }

    Answer:

    2

    Explanation:

                            above all statements form a double circular linked list;

    abc.next->next->prev->next->i

    this one points to "ghi" node the value of at particular node is 2.

    Thursday, May 24, 2007 5:41 PM
  • 77

    struct point

     {

     int x;

     int y;

     };

    struct point origin,*pp;

    main()

    {

    pp=&origin;

    printf("origin is(%d%d)\n",(*pp).x,(*pp).y);

    printf("origin is (%d%d)\n",pp->x,pp->y);

    }

               

    Answer:

    origin is(0,0)

    origin is(0,0)

    Explanation:

    pp is a pointer to structure. we can access the elements of the structure either with arrow mark or with indirection operator.

    Note:

    Since structure point  is globally declared x & y are initialized as zeroes

                           

    Thursday, May 24, 2007 5:42 PM
  • 78

    main()

    {

     int i=_l_abc(10);

                 printf("%d\n",--i);

    }

    int _l_abc(int i)

    {

     return(i++);

    }

    Answer:

    9

    Explanation:

    return(i++) it will first return i and then increments. i.e. 10 will be returned.

     

    Thursday, May 24, 2007 5:43 PM
  • 79

    main()

    {

     char *p;

     int *q;

     long *r;

     p=q=r=0;

     p++;

     q++;

     r++;

     printf("%p...%p...%p",p,q,r);

    }

    Answer:

    0001...0002...0004

    Explanation:

    ++ operator  when applied to pointers increments address according to their corresponding data-types.

    Thursday, May 24, 2007 5:43 PM
  • 80

    main()

    {

     char c=' ',x,convert(z);

     getc(c);

     if((c>='a') && (c<='z'))

     x=convert(c);

     printf("%c",x);

    }

    convert(z)

    {

      return z-32;

    }

    Answer:

    Compiler error

    Explanation:

    declaration of convert and format of getc() are wrong.

    Thursday, May 24, 2007 5:44 PM
  • 81

    main(int argc, char **argv)

    {

     printf("enter the character");

     getchar();

     sum(argv[1],argv[2]);

    }

    sum(num1,num2)

    int num1,num2;

    {

     return num1+num2;

    }

    Answer:

    Compiler error.

    Explanation:

    argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values. 

    Thursday, May 24, 2007 5:44 PM
  • 82

    # include <stdio.h>

    int one_d[]={1,2,3};

    main()

    {

     int *ptr;

     ptr=one_d;

     ptr+=3;

     printf("%d",*ptr);

    }

    Answer:

    garbage value

    Explanation:

    ptr pointer is pointing to out of the array range of one_d.

    Thursday, May 24, 2007 5:45 PM
  • 83

    # include<stdio.h>

    aaa() {

      printf("hi");

     }

    bbb(){

     printf("hello");

     }

    ccc(){

     printf("bye");

     }

    main()

    {

      int (*ptr[3])();

      ptr[0]=aaa;

      ptr[1]=bbb;

      ptr[2]=ccc;

      ptr[2]();

    }

    Answer:

    bye

    Explanation:

    ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.

    Thursday, May 24, 2007 5:45 PM
  • 84

    # include<stdio.h>

    aaa() {

      printf("hi");

     }

    bbb(){

     printf("hello");

     }

    ccc(){

     printf("bye");

     }

    main()

    {

      int (*ptr[3])();

      ptr[0]=aaa;

      ptr[1]=bbb;

      ptr[2]=ccc;

      ptr[2]();

    }

    Answer:

    bye

    Explanation:

    ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.

    Thursday, May 24, 2007 5:46 PM
  • 84

    main()

    {

     int i =0;j=0;

     if(i && j++)

                printf("%d..%d",i++,j);

    printf("%d..%d,i,j);

    }

    Answer:

    0..0

    Explanation:

    The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed.  The values of i and j remain unchanged and get printed.

         

    Thursday, May 24, 2007 5:47 PM
  • 85

    main()

    {

     int i;

     i = abc();

     printf("%d",i);

    }

    abc()

    {

     _AX = 1000;

    }

    Answer:

    1000

    Explanation:

    Normally the return value from the function is through the information from the accumulator. Here _AH is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000.

    Thursday, May 24, 2007 5:47 PM
  • 85

    int i;

                main(){

    int t;

    for ( t=4;scanf("%d",&i)-t;printf("%d\n",i))

                            printf("%d--",t--);

                            }

                // If the inputs are 0,1,2,3 find the o/p

    Answer:

                4--0

                            3--1

                            2--2    

    Explanation:

    Let us assume some x= scanf("%d",&i)-t the values during execution

                            will be,

              t        i       x

              4       0      -4

              3       1      -2

              2       2       0

    Thursday, May 24, 2007 5:48 PM
  • 87

    main(){

      int a= 0;int b = 20;char x =1;char y =10;

      if(a,b,x,y)

            printf("hello");

     }

    Answer:

    hello

    Explanation:

    The comma operator has associativity from left to right. Only the rightmost value is returned and the other values are evaluated and ignored. Thus the value of last variable y is returned to check in if. Since it is a non zero value if becomes true so, "hello" will be printed.

     

    Thursday, May 24, 2007 5:48 PM
  • 88

    main(){

     unsigned int i;

     for(i=1;i>-2;i--)

                            printf("c aptitude");

    }

    Explanation:

    i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop.

    Thursday, May 24, 2007 5:49 PM
  • 89

    In the following pgm add a  stmt in the function  fun such that the address of

    'a' gets stored in 'j'.

    main(){

      int * j;

      void fun(int **);

      fun(&j);

     }

     void fun(int **k) {

      int a =0;

      /* add a stmt here*/

     }

    Answer:

                            *k = &a

    Explanation:

                            The argument of the function is a pointer to a pointer.

    Thursday, May 24, 2007 5:49 PM
  • 90

    What are the following notations of defining functions known as?

    i.      int abc(int a,float b)

                            {

                            /* some code */

     }

    ii.    int abc(a,b)

            int a; float b;

                            {

                            /* some code*/

                            }

    Answer:

    i.  ANSI C notation

    ii. Kernighan & Ritche notation

     

    Thursday, May 24, 2007 5:50 PM
  • 91

    main()

    {

    char *p;

    p="%d\n";

               p++;

               p++;

               printf(p-2,300);

    }

    Answer:

                300

    Explanation:

    The pointer points to % since it is incremented twice and again decremented by 2, it points to '%d\n' and 300 is printed.

    Thursday, May 24, 2007 5:51 PM
  • 92

    main(){

     char a[100];

     a[0]='a';a[1]]='b';a[2]='c';a[4]='d';

     abc(a);

    }

    abc(char a[]){

     a++;

                 printf("%c",*a);

     a++;

     printf("%c",*a);

    }

    Explanation:

    The base address is modified only in function and as a result a points to 'b' then after incrementing to 'c' so bc will be printed.

    Thursday, May 24, 2007 5:51 PM
  • 93

     

    func(a,b)

    int a,b;

    {

     return( a= (a==b) );

    }

    main()

    {

    int process(),func();

    printf("The value of process is %d !\n ",process(func,3,6));

    }

    process(pf,val1,val2)

    int (*pf) ();

    int val1,val2;

    {

    return((*pf) (val1,val2));

     }

    Answer:

    The value if process is 0 !

    Explanation:

    The function 'process' has 3 parameters - 1, a pointer to another function  2 and 3, integers. When this function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the operation performed by the function 'func'. The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in turn is returned by the function 'process'.

     

    Thursday, May 24, 2007 5:52 PM
  • 94

    void main()

    {

                static int i=5;

                if(--i){

                            main();

                            printf("%d ",i);

                }

    }

    Answer:

     0 0 0 0

    Explanation:

                The variable "I" is declared as static, hence memory for I will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless I becomes equal to 0, and since main() is recursively called, so the value of static I ie., 0 will be printed every time the control is returned.

    Thursday, May 24, 2007 5:53 PM
  • 95

    void main()

    {

                int k=ret(sizeof(float));

                printf("\n here value is %d",++k);

    }

    int ret(int ret)

    {

                ret += 2.5;

                return(ret);

    }

    Answer:

     Here value is 7

    Explanation:

                The int ret(int ret), ie., the function name and the argument name can be the same.

                Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed,  after the first expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit type conversion from float to int. The ret is returned in main() it is printed after and preincrement.

    Thursday, May 24, 2007 5:54 PM
  • 96

    void main()

    {

                char a[]="12345\0";

                int i=strlen(a);

                printf("here in 3 %d\n",++i);

    }

    Answer:

    here in 3 6

    Explanation:

                The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'I' will hold the value equal to 5, after the pre-increment in the printf statement, the 6 will be printed.

    Thursday, May 24, 2007 5:55 PM
  • 97

    void main()

    {

                unsigned giveit=-1;

                int gotit;

                printf("%u ",++giveit);

                printf("%u \n",gotit=--giveit);

    }

    Answer:

     0 65535

    Explanation:

    Thursday, May 24, 2007 5:56 PM
  • 98          

      void main()

    {

                int i;

                char a[]="\0";

                if(printf("%s\n",a))

                            printf("Ok here \n");

                else

                            printf("Forget it\n");

    }

    Answer:

     Ok here

    Explanation:

    Printf will return how many characters does it print. Hence printing a null character returns 1 which makes the if statement true, thus "Ok here" is printed.

    Thursday, May 24, 2007 5:57 PM
  • 99

    void main()

    {

                void *v;

                int integer=2;

                int *i=&integer;

                v=i;

                printf("%d",(int*)*v);

    }

    Answer:

    Compiler Error. We cannot apply indirection on type void*.

    Explanation:

    Void pointer is a generic pointer type. No pointer arithmetic can be done on it. Void pointers are normally used for,

    1.      Passing generic pointers to functions and returning such pointers.

    2.      As a intermediate pointer type.

    3.      Used when the exact pointer type will be known at a later point of time.

    Thursday, May 24, 2007 5:58 PM
  • 100

     

    void main()

    {

                int i=i++,j=j++,k=k++;

    printf(“%d%d%d”,i,j,k);

    }

    Answer:

    Garbage values.

    Explanation:

    An identifier is available to use in program code from the point of its declaration.

    So expressions such as  i = i++ are valid statements. The i, j and k are automatic variables and so they contain some garbage value. Garbage in is garbage out (GIGO).

    Thursday, May 24, 2007 5:58 PM
  • 101

    void main()

    {

                void *v;

                int integer=2;

                int *i=&integer;

                v=i;

                printf("%d",(int*)*v);

    }

    Answer:

    Compiler Error. We cannot apply indirection on type void*.

    Explanation:

    Void pointer is a generic pointer type. No pointer arithmetic can be done on it. Void pointers are normally used for,

    1.      Passing generic pointers to functions and returning such pointers.

    2.      As a intermediate pointer type.

    3.      Used when the exact pointer type will be known at a later point of time.

    Friday, May 25, 2007 5:42 AM
  • 102

    void main()

    {

                int i=i++,j=j++,k=k++;

    printf(“%d%d%d”,i,j,k);

    }

    Answer:

    Garbage values.

    Explanation:

    An identifier is available to use in program code from the point of its declaration.

    So expressions such as  i = i++ are valid statements. The i, j and k are automatic variables and so they contain some garbage value. Garbage in is garbage out (GIGO).

    Friday, May 25, 2007 5:42 AM
  • 103

    void main()

    {

                static int i=i++, j=j++, k=k++;

    printf(“i = %d j = %d k = %d”, i, j, k);

    }

    Answer:

    i = 1 j = 1 k = 1

    Explanation:

    Since static variables are initialized to zero by default.

    Friday, May 25, 2007 5:43 AM
  • 104

    void main()

    {

                while(1){

                            if(printf("%d",printf("%d")))

                                        break;

                            else

                                        continue;

                }

    }

    Answer:

    Garbage values

    Explanation:

    The inner printf executes first to print some garbage value. The printf returns no of characters printed and this value also cannot be predicted. Still the outer printf  prints something and so returns a non-zero value. So it encounters the break statement and comes out of the while statement.

    Friday, May 25, 2007 5:44 AM
  • 105

    main()

    {

                unsigned int i=10;

                while(i-->=0)

                            printf("%u ",i);

     

    }

    Answer:

                10 9 8 7 6 5 4 3 2 1 0 65535 65534…..

    Explanation:

    Since i is an unsigned integer it can never become negative. So the expression i-- >=0  will always be true, leading to an infinite loop.         

    Friday, May 25, 2007 5:45 AM
  • 106

    #include<conio.h>

    main()

    {

                int x,y=2,z,a;

                if(x=y%2) z=2;

                a=2;

                printf("%d %d ",z,x);

    }

     Answer:

    Garbage-value 0

    Explanation:

    The value of y%2 is 0. This value is assigned to x. The condition reduces to if (x) or in other words if(0) and so z goes uninitialized.

    Thumb Rule: Check all control paths to write bug free code.

    Friday, May 25, 2007 5:45 AM
  • 107

    main()

    {

                int a[10];

                printf("%d",*a+1-*a+3);

    }

    Answer:

    4 

    Explanation:

                *a and -*a cancels out. The result is as simple as 1 + 3 = 4 !    

    Friday, May 25, 2007 5:46 AM
  • 108

    #define prod(a,b) a*b

    main()

    {

                int x=3,y=4;

                printf("%d",prod(x+2,y-1));

    }

    Answer:

    10

    Explanation:

                The macro expands and evaluates to as:

                x+2*y-1 => x+(2*y)-1 => 10

    Friday, May 25, 2007 5:46 AM
  • 109

    main()

    {

                unsigned int i=65000;

                while(i++!=0);

                printf("%d",i);

    }

    Answer:

     1

    Explanation:

    Note the semicolon after the while statement. When the value of i becomes 0 it comes out of while loop. Due to post-increment on i the value of i while printing is 1.

     

    Friday, May 25, 2007 5:47 AM
  • 110

    main()

    {

                int i=0;

                while(+(+i--)!=0)

                            i-=i++;

                printf("%d",i);

    }

    Answer:

    -1

    Explanation:

    Unary + is the only dummy operator in C. So it has no effect on the expression and now the while loop is,         while(i--!=0) which is false and so breaks out of while loop. The value –1 is printed due to the post-decrement operator.

    Friday, May 25, 2007 5:48 AM
  • 111

    main()

    {

                float f=5,g=10;

                enum{i=10,j=20,k=50};

                printf("%d\n",++k);

                printf("%f\n",f<<2);

                printf("%lf\n",f%g);

                printf("%lf\n",fmod(f,g));

    }

    Answer:

    Line no 5: Error: Lvalue required

    Line no 6: Cannot apply leftshift to float

    Line no 7: Cannot apply mod to float

    Explanation:

                            Enumeration constants cannot be modified, so you cannot apply ++.

                            Bit-wise operators and % operators cannot be applied on float values.

                            fmod() is to find the modulus values for floats as % operator is for ints. 

    Friday, May 25, 2007 5:48 AM
  • 112

    main()

    {

                int i=10;

                void pascal f(int,int,int);

    f(i++,i++,i++);

                printf(" %d",i);

    }

    void pascal f(integer :i,integer:j,integer :k)

    {

    write(i,j,k);

    }

    Answer:

    Compiler error:  unknown type integer

    Compiler error:  undeclared function write

    Explanation:

    Pascal keyword doesn’t mean that pascal code can be used. It means that the function follows Pascal argument passing mechanism in calling the functions.

    Friday, May 25, 2007 5:49 AM
  • 113

    void pascal f(int i,int j,int k)

    {

    printf(“%d %d %d”,i, j, k);

    }

    void cdecl f(int i,int j,int k)

    {

    printf(“%d %d %d”,i, j, k);

    }

    main()

    {

                int i=10;

    f(i++,i++,i++);

                printf(" %d\n",i);

    i=10;

    f(i++,i++,i++);

    printf(" %d",i);

    }

    Answer:

    10 11 12 13

    12 11 10 13

    Explanation:

    Pascal argument passing mechanism forces the arguments to be called from left to right. cdecl is the normal C argument passing mechanism where the arguments are passed from right to left.

    Friday, May 25, 2007 5:49 AM
  • 114

    What is the output of the program given below

     

    main()

        {

           signed char i=0;

           for(;i>=0;i++) ;

           printf("%d\n",i);

        }

    Answer

                            -128

    Explanation

    Notice the semicolon at the end of the for loop. THe initial value of the i is set to 0. The inner loop executes to increment the value from 0 to 127 (the positive range of char) and then it rotates to the negative value of -128. The condition in the for loop fails and so comes out of the for loop. It prints the current value of i that is -128.

    Friday, May 25, 2007 5:50 AM
  • 115

    main()

        {

           unsigned char i=0;

           for(;i>=0;i++) ;

           printf("%d\n",i);

        }

    Answer

                infinite loop

    Explanation

    The difference between the previous question and this one is that the char is declared to be unsigned. So the i++ can never yield negative value and i>=0 never becomes false so that it can come out of the for loop.

    Friday, May 25, 2007 5:51 AM
  • 116

    main()

                {

           char i=0;

           for(;i>=0;i++) ;

           printf("%d\n",i);

           

     }

    Answer:

                            Behavior is implementation dependent.

    Explanation:

    The detail if the char is signed/unsigned by default is implementation dependent. If the implementation treats the char to be signed by default the program will print –128 and terminate. On the other hand if it considers char to be unsigned by default, it goes to infinite loop.

    Rule:

    You can write programs that have implementation dependent behavior. But dont write programs that depend on such behavior.

    Friday, May 25, 2007 5:52 AM
  • 117

    Is the following statement a declaration/definition. Find what does it mean?

    int (*x)[10];

    Answer

                            Definition.

                x is a pointer to array of(size 10) integers.

     

                            Apply clock-wise rule to find the meaning of this definition.

    Friday, May 25, 2007 5:54 AM
  • 118 What is the output for the program given below

     

         typedef enum errorType{warning, error, exception,}error;

         main()

        {

            error g1;

            g1=1;

            printf("%d",g1);

         }

    Answer

                            Compiler error: Multiple declaration for error

    Explanation

    The name error is used in the two meanings. One means that it is a enumerator constant with value 1. The another use is that it is a type name (due to typedef) for enum errorType. Given a situation the compiler cannot distinguish the meaning of error to know in what sense the error is used:

                error g1;

    g1=error;

                // which error it refers in each case?

    When the compiler can distinguish between usages then it will not issue error (in pure technical terms, names can only be overloaded in different namespaces).

    Note: the extra comma in the declaration,

    enum errorType{warning, error, exception,}

    is not an error. An extra comma is valid and is provided just for programmer’s convenience.

    Friday, May 25, 2007 5:55 AM
  • 119    

    typedef struct error{int warning, error, exception;}error;

         main()

        {

            error g1;

            g1.error =1;

            printf("%d",g1.error);

         }

     

    Answer

                            1

    Explanation

    The three usages of name errors can be distinguishable by the compiler at any instance, so valid (they are in different namespaces).

    Typedef struct error{int warning, error, exception;}error;

    This error can be used only by preceding the error by struct kayword as in:

    struct error someError;

    typedef struct error{int warning, error, exception;}error;

    This can be used only after . (dot) or -> (arrow) operator preceded by the variable name as in :

    g1.error =1;

                printf("%d",g1.error);

                            typedef struct error{int warning, error, exception;}error;

    This can be used to define variables without using the preceding struct keyword as in:

    error g1;

    Since the compiler can perfectly distinguish between these three usages, it is perfectly legal and valid.

     

    Note

    This code is given here to just explain the concept behind. In real programming don’t use such overloading of names. It reduces the readability of the code. Possible doesn’t mean that we should use it!

    Friday, May 25, 2007 6:01 AM
  • 120

    #ifdef something

    int some=0;

    #endif

     

    main()

    {

    int thing = 0;

    printf("%d %d\n", some ,thing);

    }

     

    Answer:

                            Compiler error : undefined symbol some

    Explanation:

    This is a very simple example for conditional compilation. The name something is not already known to the compiler making the declaration

    int some = 0;

    effectively removed from the source code.

     

    Friday, May 25, 2007 6:01 AM
  • why don't you just post these questions for others to reply?
    after all forums are meant to share, learn & grow.
    though yo are providing info but these won't make ppl here to scratch their mind because they are already getting answers.

    QUESTION:
    i want to print the numbrs from 1 to N(user input) without using any looping techinque or goto or recursion or indirect recursion.
    post the answer
    Friday, May 25, 2007 12:20 PM
  • Can u give me ur yahoo ID
    Friday, May 25, 2007 2:59 PM
  • 122

     int swap(int *a,int *b)

    {

     *a=*a+*b;*b=*a-*b;*a=*a-*b;

    }

    main()

    {

                            int x=10,y=20;

                swap(&x,&y);

                            printf("x= %d y = %d\n",x,y);

    }

    Answer

                x = 20 y = 10

    Explanation

    This is one way of swapping two values. Simple checking will help understand this.

    Friday, May 25, 2007 2:59 PM
  • 123

     main()

    {         

    char *p = “ayqm”;

    printf(“%c”,++*(p++));

    }

    Answer:

    b         

    Friday, May 25, 2007 3:00 PM
  • 124

    main()

                {

                 int i=5;

                 printf("%d",++i++);

    }

    Answer:

                            Compiler error: Lvalue required in function main

    Explanation:

                            ++i yields an rvalue.  For postfix ++ to operate an lvalue is required.

    Friday, May 25, 2007 3:00 PM
  • #include <stdio.h>
    int main(void)

    {
                   char input[100];
                   scanf("%100s", input);
                   printf("1 to %s\n", input);
                   return 0;
    }

    Friday, May 25, 2007 3:08 PM
  • 127

    main()

    {

    char *p = “ayqm”;

    char c;

    c = ++*p++;

    printf(“%c”,c);

    }

    Answer:

    b

    Explanation:

    There is no difference between the expression ++*(p++) and ++*p++. Parenthesis just works as a visual clue for the reader to see which expression is first evaluated.

    Friday, May 25, 2007 3:11 PM
  • 128

    int aaa() {printf(“Hi”);}

    int bbb(){printf(“hello”);}

    iny ccc(){printf(“bye”);}

     

    main()

    {

    int ( * ptr[3]) ();

    ptr[0] = aaa;

    ptr[1] = bbb;

    ptr[2] =ccc;

    ptr[2]();

    }

    Answer:

     bye

    Explanation:

    int (* ptr[3])() says that ptr is an array of pointers to functions that takes no arguments and returns the type int. By the assignment ptr[0] = aaa; it means that the first function pointer in the array is initialized with the address of the function aaa. Similarly, the other two array elements also get initialized with the addresses of the functions bbb and ccc. Since ptr[2] contains the address of the function ccc, the call to the function ptr[2]() is same as calling ccc(). So it results in printing  "bye".

    Friday, May 25, 2007 3:12 PM
  • hey you still are continuing with it.
    what's the matter/
    & why do you want my id?
    i'm smelling a rat
    Friday, May 25, 2007 4:24 PM
  • i think answer should be "z".
    don't give wrong answers
    Sunday, May 27, 2007 12:51 PM
  • 129

    char *someFun1()

                {

                char temp[ ] = “string";

                return temp;

                }

                char *someFun2()

                {

                char temp[ ] = {‘s’, ‘t’,’r’,’i’,’n’,’g’};

                return temp;

                }

                int main()

                {

                puts(someFun1());

                puts(someFun2());

                }

    Answer:

                Garbage values.

    Explanation:

                Both the functions suffer from the problem of dangling pointers. In someFun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string “string”. This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function someFun2() also suffers from the same problem but the problem can be easily identified in this case.

    Monday, May 28, 2007 7:58 AM
  • @sunil

     

    Coz i want to share my knowledge with u ........

    Monday, May 28, 2007 7:59 AM
  • 130

    char *someFun()

                {

                char *temp = “string constant";

                return temp;

                }

                int main()

                {

                puts(someFun());

                }

    Answer:

                string constant

    Explanation:

                The program suffers no problem and gives the output correctly because the character constants are stored in code/data area and not allocated in stack, so this doesn’t lead to dangling pointers.

    Monday, May 28, 2007 8:00 AM
  •  

    131

    Printf can be implemented by using  __________ list.

    Answer:

                            Variable length argument lists

    Monday, May 28, 2007 8:00 AM
  • 132

    main()

                            {

                                        extern int i;

                            {          int i=20;

                             {        

                               const volatile unsigned i=30; printf("%d",i);

                             }

                            printf("%d",i);

                            }

                              printf("%d",i);

                }         

                int i;

    Monday, May 28, 2007 8:01 AM
  • 133

    main()

    {

                            int a=10,*j;

                void *k;

                            j=k=&a;

                j++; 

                            k++;

                printf("\n %u %u ",j,k);

    }

    Answer:

                            Compiler error: Cannot increment a void pointer

    Explanation:

    Void pointers are generic pointers and they can be used only when the type is not known and as an intermediate address storage type. No pointer arithmetic can be done on it and you cannot apply indirection operator (*) on void pointers.

    Monday, May 28, 2007 8:02 AM
  • 134

    main()

    {         

    char a[4]="HELL";

    printf("%s",a);

    }

    Answer:

                            HELL%@!~@!@???@~~!

    Explanation:

    The character array has the memory just enough to hold the string “HELL” and doesnt have enough space to store the terminating null character. So it prints the HELL correctly and continues to print garbage values till it          accidentally comes across a NULL character.

    Monday, May 28, 2007 8:02 AM
  • 135

    main()

    {

    char a[4]="HELLO";

    printf("%s",a);

    }         

    Answer:

                            Compiler error: Too many initializers

    Explanation:

    The array a is of size 4 but the string constant requires 6 bytes to get stored.

    Monday, May 28, 2007 8:03 AM
  • 136   

     

             Is this code legal?

    int *ptr;

    ptr = (int *) 0x400;

    Answer:

                            Yes

    Explanation:

    The pointer ptr will point at the integer in the memory location 0x400.

    Monday, May 28, 2007 8:04 AM
  • 137

    void main()

    {

    char ch;

    for(ch=0;ch<=127;ch++)

    printf(“%c   %d \n“, ch, ch);

    }

    Answer:

                Implementaion dependent

    Explanation:

    The char type may be signed or unsigned by default. If it is signed then ch++ is executed after ch reaches 127 and rotates back to -128. Thus ch is always smaller than 127.

    Monday, May 28, 2007 8:04 AM
  • 138

    void main()

    {

    int i=10, j=2;

    int *ip= &i, *jp = &j;

    int k = *ip/*jp;

    printf(“%d”,k);

    }         

    Answer:

    Compiler Error: “Unexpected end of file in comment started in line 5”.

    Explanation:

    The programmer intended to divide two integers, but by the “maximum munch” rule, the compiler treats the operator sequence / and * as /* which happens to be the starting of comment. To force what is intended by the programmer,

    int k = *ip/ *jp;

    // give space explicity separating / and *

    //or

    int k = *ip/(*jp);

    // put braces to force the intention 

    will solve the problem. 

     

    Monday, May 28, 2007 8:05 AM
  • 139

    Which version do you prefer of the following two,

    1) printf(“%s”,str);        // or the more curt one

    2) printf(str);

    Answer & Explanation:

    Prefer the first one. If the str contains any  format characters like %d then it will result in a subtle bug.

    Monday, May 28, 2007 8:06 AM
  • 139

    char inputString[100] = {0};

    To get string input from the keyboard which one of the following is better?

                1) gets(inputString)

                2) fgets(inputString, sizeof(inputString), fp)

    Answer & Explanation:

    The second one is better because gets(inputString) doesn't know the size of the string passed and so, if a very big input (here, more than 100 chars) the charactes will be written past the input string. When fgets is used with stdin performs the same operation as gets but is safe.

     

    Monday, May 28, 2007 8:06 AM
  • 140

    void main()

    {

    printf(“sizeof (void *) = %d \n“, sizeof( void *));

                printf(“sizeof (int *)    = %d \n”, sizeof(int *));

                printf(“sizeof (double *)  = %d \n”, sizeof(double *));

                printf(“sizeof(struct unknown *) = %d \n”, sizeof(struct unknown *));

                }

    Answer            :

    sizeof (void *) = 2

    sizeof (int *)    = 2

    sizeof (double *)  =  2

    sizeof(struct unknown *) =  2

    Explanation:

    The pointer to any type is of same size.

    Monday, May 28, 2007 8:07 AM
  • 141

     

    Is the following code legal?

    void main()

    {

    typedef struct a aType;

    aType someVariable;

    struct a

    {

    int x;

          aType *b;

                  };

    }

    Answer:

                            No

    Explanation:

                            When the declaration,

    typedef struct a aType;

    is encountered body of struct a is not known. This is known as ‘incomplete types’.

    Monday, May 28, 2007 8:08 AM
  • 142

    Is the following code legal?

    typedef struct a aType;

    struct a

    {

    int x;

    aType *b;

    };

    Answer:

                Yes

    Explanation:

    The typename aType is known at the point of declaring the structure, because it is already typedefined.

    Monday, May 28, 2007 8:08 AM
  • 143

    Is the following code legal?

    typedef struct a

        {

    int x;

     aType *b;

        }aType

    Answer:

                            No

    Explanation:

    The typename aType is not known at the point of declaring the structure (forward references are not made for typedefs).

    Monday, May 28, 2007 8:09 AM
  • 144

    Is the following code legal?

    struct a

        {

    int x;

                struct a *b;

        }

    Answer:

    Yes.

    Explanation:

    *b is a pointer to type struct a and so is legal. The compiler knows, the size of the pointer to a structure even before the size of the structure

    is determined(as you know the pointer to any type is of same size). This type of structures is known as ‘self-referencing’ structure.

     

    Monday, May 28, 2007 8:10 AM
  • 145        

        Is the following code legal?

    struct a

        {

    int x;

     struct a b;

        }

    Answer:

                            No

    Explanation:

    Is it not legal for a structure to contain a member that is of the same

    type as in this case. Because this will cause the structure declaration to be recursive without end.

    Monday, May 28, 2007 8:10 AM
  • 146

     

    #define assert(cond) if(!(cond)) \

      (fprintf(stderr, "assertion failed: %s, file %s, line %d \n",#cond,\

     __FILE__,__LINE__), abort())

     

    void main()

    {

    int i = 10;

    if(i==0)

        assert(i < 100);

    else

        printf("This statement becomes else for if in assert macro");

    }

    Answer:

    No output

    Explanation:

    The else part in which the printf is there becomes the else for if in the assert macro. Hence nothing is printed.

    The solution is to use conditional operator instead of if statement,

    #define assert(cond) ((cond)?(0): (fprintf (stderr, "assertion failed: \ %s, file %s, line %d \n",#cond, __FILE__,__LINE__), abort()))

     

    Note:

    However this problem of “matching with nearest else” cannot be solved by the usual method of placing the if statement inside a block like this,

    #define assert(cond) { \

    if(!(cond)) \

      (fprintf(stderr, "assertion failed: %s, file %s, line %d \n",#cond,\

     __FILE__,__LINE__), abort()) \

    }

    Monday, May 28, 2007 8:11 AM

All replies

  • 1.      void main()

    {

                int  const * p=5;

                printf("%d",++(*p));

    }

    Answer:

                            Compiler error: Cannot modify a constant value.

    Explanation:   

    p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

    Thursday, May 24, 2007 1:29 PM
  • 1.      main()

    {

                char s[ ]="man";

                int i;

                for(i=0;s[ i ];i++)

                printf("\n%c%c%c%c",s[ i ],*(s+i),*(i+s),iSleep);

    }

    Answer:

                            mmmm

                           aaaa

                           nnnn

    Explanation:

    sIdea, *(i+s), *(s+i), iSleep are all different ways of expressing the same idea. Generally  array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as sIdea. iSleep may be surprising. But in the  case of  C  it is same as sIdea.

    Thursday, May 24, 2007 1:32 PM
  • 1.      main()

    {

                float me = 1.1;

                double you = 1.1;

                if(me==you)

    printf("I love U");

    else

                            printf("I hate U");

    }

    Answer:

    I hate U

    Explanation:

    For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value  represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.

     

    Thursday, May 24, 2007 1:34 PM
  • 1.      main()

                {

                static int var = 5;

                printf("%d ",var--);

                if(var)

                            main();

                }

    Answer:

    5 4 3 2 1

                Explanation:

    When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively. 

    Thursday, May 24, 2007 1:35 PM
  • 5.      main()

    {

                 int c[ ]={2.8,3.4,4,6.7,5};

                 int j,*p=c,*q=c;

                 for(j=0;j<5;j++) {

                            printf(" %d ",*c);

                            ++q;     }

                 for(j=0;j<5;j++){

    printf(" %d ",*p);

    ++p;     }

    }

     

    Answer:

                            2 2 2 2 2 2 3 4 6 5

                Explanation:

    Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

    Thursday, May 24, 2007 1:36 PM
  • 6.      main()

    {

                extern int i;

                i=20;

    printf("%d",i);

    }

     

    Answer: 

    Linker Error : Undefined symbol '_i'

    Explanation:

                            extern storage class in the following declaration,

                                        extern int i;

    specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

     

    Thursday, May 24, 2007 1:37 PM
  • 7.      main()

    {

                int i=-1,j=-1,k=0,l=2,m;

                m=i++&&j++&&k++||l++;

                printf("%d %d %d %d %d",i,j,k,l,m);

    }

    Answer:

                            0 0 1 3 1

    Explanation :

    Logical operations always give a result of 1 or 0 . And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression  i++ && j++ && k++’ is executed first. The result of this expression is 0    (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

    Thursday, May 24, 2007 1:38 PM
  • 8.      main()

    {

                char *p;

                printf("%d %d ",sizeof(*p),sizeof(p));

    }

     

    Answer:

                            1 2

    Explanation:

    The sizeof() operator gives the number of bytes taken by its operand. P is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.

    Thursday, May 24, 2007 1:39 PM
  • 9.      main()

    {

                int i=3;

                switch(i)

                 {

                    defaultStick out tonguerintf("zero");

                    case 1: printf("one");

                               break;

                   case 2Stick out tonguerintf("two");

                              break;

                  case 3: printf("three");

                              break;

                  } 

    }

    Answer :

    three

    Explanation :

    The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

    Thursday, May 24, 2007 1:40 PM
  • cool, start at beginners level questions and then end with expert level questions, it would be good to have t...
    Thursday, May 24, 2007 1:40 PM
  • 10.      main()

    {

                  printf("%x",-1<<4);

    }

    Answer:

    fff0

    Explanation :

    -1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

    Thursday, May 24, 2007 1:41 PM
  • 11.      main()

    {

                  printf("%x",-1<<4);

    }

    Answer:

    fff0

    Explanation :

    -1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.

    Thursday, May 24, 2007 1:42 PM
  • 11.      main()

    {

                char string[]="Hello World";

                display(string);

    }

    void display(char *string)

    {

                printf("%s",string);

    }

                Answer:

    Compiler Error : Type mismatch in redeclaration of function display

                Explanation :

    In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and

    return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.

    Thursday, May 24, 2007 3:30 PM
  • 12.      main()

    {

                int c=- -2;

                printf("c=%d",c);

    }

    Answer:

                                        c=2;

                Explanation:

    Here unary minus (or negation) operator is used twice. Same maths  rules applies, ie. minus * minus= plus.

    Note:

    However you cannot give like --2. Because -- operator can  only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

    Thursday, May 24, 2007 3:30 PM
  • 13.      #define int char

    main()

    {

                int i=65;

                printf("sizeof(i)=%d",sizeof(i));

    }

    Answer:

                            sizeof(i)=1

    Explanation:

    Since the #define replaces the string  int by the macro char

    Thursday, May 24, 2007 3:31 PM
  • 14.      main()

    {

    int i=10;

    i=!i>14;

    Printf ("i=%d",i);

    }

    Answer:

    i=0

     

     

                Explanation:

    In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’ symbol.  ! is a unary logical operator. !i (!10) is 0 (not of true is false).  0>14 is false (zero).

    Thursday, May 24, 2007 3:32 PM
  • 15.      #include<stdio.h>

    main()

    {

    char s[]={'a','b','c','\n','c','\0'};

    char *p,*str,*str1;

    p=&s[3];

    str=p;

    str1=s;

    printf("%d",++*p + ++*str1-32);

    }

    Answer:

    77       

    Explanation:

    p is pointing to character '\n'. str1 is pointing to character 'a' ++*p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.

     Now performing (11 + 98 – 32), we get 77("M");

     So we get the output 77 :: "M" (Ascii is 77).

    Thursday, May 24, 2007 3:32 PM
  • 16.      #include<stdio.h>

    main()

    {

    int a[2][2][2] = { {10,2,3,4}, {5,6,7,8}  };

    int *p,*q;

    p=&a[2][2][2];

    *q=***a;

    printf("%d----%d",*p,*q);

    }

    Answer:

    SomeGarbageValue---1

    Explanation:

    p=&a[2][2][2]  you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.

               

    Thursday, May 24, 2007 3:33 PM
  • 17.      #include<stdio.h>

    main()

    {

    struct xx

    {

          int x=3;

          char name[]="hello";

     };

    struct xx *s;

    printf("%d",s->x);

    printf("%s",s->name);

    }

                Answer:

    Compiler Error

    Explanation:

    You should not initialize variables in declaration

     

    Thursday, May 24, 2007 3:33 PM
  • 18.      #include<stdio.h>

    main()

    {

    struct xx

    {

    int x;

    struct yy

    {

    char s;

                struct xx *p;

    };

    struct yy *q;

    };

    }

    Answer:

    Compiler Error

    Explanation:

    The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.

    Thursday, May 24, 2007 3:34 PM
  • 19.      main()

    {

    printf("\nab");

    printf("\bsi");

    printf("\rha");

    }

    Answer:

    hai

    Explanation:

    \n  - newline

    \b  - backspace

    \r  - linefeed

    Thursday, May 24, 2007 3:34 PM
  • 20.      main()

    {

    int i=5;

    printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);

    }

    Answer:

    45545

    Explanation:

    The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack. and the  evaluation is from right to left, hence the result.

    Thursday, May 24, 2007 3:35 PM
  • 21.      #define square(x) x*x

    main()

    {

    int i;

    i = 64/square(4);

    printf("%d",i);

    }

    Answer:

    64

    Explanation:

    the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

     

    Thursday, May 24, 2007 3:35 PM
  • 22.      main()

    {

    char *p="hai friends",*p1;

    p1=p;

    while(*p!='\0') ++*p++;

    printf("%s   %s",p,p1);

    }

    Answer:

    ibj!gsjfoet

                Explanation:

                            ++*p++ will be parse in the given order

    Ø      *p that is value at the location currently pointed by p will be taken

    Ø      ++*p the retrieved value will be incremented

    Ø      when ; is encountered the location will be incremented that is p++ will be executed

    Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1doesnot print anything
    Thursday, May 24, 2007 3:36 PM
  • 23.      #include <stdio.h>

    #define a 10

    main()

    {

    #define a 50

    printf("%d",a);

    }

    Answer:

    50

    Explanation:

    The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

    Thursday, May 24, 2007 3:37 PM
  • 1.      #define clrscr() 100

    main()

    {

    clrscr();

    printf("%d\n",clrscr());

    }

    Answer:

    100

    Explanation:

    Preprocessor executes as a seperate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input  program to compiler looks like this :

                            main()

                            {

                                 100;

                                 printf("%d\n",100);

                            }

                Note:  

    100; is an executable statement but with no action. So it doesn't give any problem

    Thursday, May 24, 2007 3:38 PM
  • Question