c aptitude

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  Predict the output or error(s) for the following:

   

        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, June 14, 2007 12:39 PM
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   main()

  {

              char s[ ]="man";

              int i;

              for(i=0;s[ i ];i++)

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

  }

  Answer:

                          mmmm

                         aaaa

                         nnnn

  Explanation:

  s, *(i+s), *(s+i), i 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 s. i may be surprising. But in the  case of  C  it is same as s.

   

  Thursday, June 14, 2007 12:39 PM
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  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.

  Rule of Thumb:

  Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ) . 

  Thursday, June 14, 2007 12:40 PM
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  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, June 14, 2007 12:40 PM
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  1.      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, June 14, 2007 12:41 PM
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  1.      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, June 14, 2007 12:42 PM
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  1.      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, June 14, 2007 12:42 PM
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  1.      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, June 14, 2007 12:43 PM
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  1.      main()

  {

              int i=3;

              switch(i)

               {

                  defaultrintf("zero");

                  case 1: printf("one");

                             break;

                 case 2rintf("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, June 14, 2007 12:43 PM
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  1.      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.

   

  2.      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, June 14, 2007 12:43 PM
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  1.      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, June 14, 2007 12:44 PM
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  1.      #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, June 14, 2007 12:44 PM
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  @Senthil - Nice work.. it would be still better if you will post the questions alone and we will try the answers...

  Thursday, June 14, 2007 1:45 PM
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  I agree with you Raghuram, it should be interactive.
  

  Thursday, June 14, 2007 2:46 PM
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  Yes make it interactive and please use the code block to put the code here...
  

  Friday, June 15, 2007 4:14 AM
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  thank u for your response .. here after i ll post only questions that are asked in campus time....
  
  
  and also u try to post some important questions......
  

  Friday, June 15, 2007 9:07 AM
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  <!--[if !supportLists]-->1.      <!--[endif]-->main()

  {

  int i=10;

  i=!i>14;

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

  }

  Friday, June 15, 2007 9:07 AM
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  <!--[if !supportLists]-->1.      <!--[endif]-->#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);

  }

  Friday, June 15, 2007 9:07 AM
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  <!--[if !supportLists]-->1.      <!--[endif]-->#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.

  Friday, June 15, 2007 9:08 AM
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  <!--[if !supportLists]-->1.      <!--[endif]-->#include<stdio.h>

  main()

  {

  struct xx

  {

        int x=3;

        char name[]="hello";

   };

  struct xx *s;

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

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

  }

  Friday, June 15, 2007 9:08 AM
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  <!--[if !supportLists]-->1.      <!--[endif]-->#include<stdio.h>

  main()

  {

  struct xx

  {

  int x;

  struct yy

  {

  char s;

              struct xx *p;

  };

  struct yy *q;

  };

  }

  Friday, June 15, 2007 9:08 AM
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  <!--[if !supportLists]-->1.      <!--[endif]-->main()

  {

  printf("\nab");

  printf("\bsi");

  printf("\rha");

  }

  Answer:

  hai

  Explanation:

  \n  - newline

  \b  - backspace

  \r  - linefeed

  Friday, June 15, 2007 9:09 AM
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  <!--[if !supportLists]-->1.      <!--[endif]-->main()

  {

  int i=5;

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

  }

  Friday, June 15, 2007 9:09 AM
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  <!--[if !supportLists]-->1.      <!--[endif]-->#define square(x) x*x

  main()

  {

  int i;

  i = 64/square(4);

  printf("%d",i);

  }

  Friday, June 15, 2007 9:09 AM
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  <!--[if !supportLists]-->1.      <!--[endif]-->main()

  {

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

  p1=p;

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

  printf("%s   %s",p,p1);

  }

  Friday, June 15, 2007 9:09 AM
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  <!--[if !supportLists]-->1.      <!--[endif]-->#include <stdio.h>

  #define a 10

  main()

  {

  #define a 50

  printf("%d",a);

  }

  Friday, June 15, 2007 9:10 AM
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  <!--[if !supportLists]-->1.      <!--[endif]-->#define clrscr() 100

  main()

  {

  clrscr();

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

  }

  Friday, June 15, 2007 9:10 AM
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  <!--[if !supportLists]-->1.      <!--[endif]-->main()

  {

  printf("%p",main);

  }

  Friday, June 15, 2007 9:11 AM
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  main()

  {

  clrscr();

  }

  clrscr();

  Friday, June 15, 2007 9:11 AM
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  enum colors {BLACK,BLUE,GREEN}

   main()

  {

   

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

    

   return(1);

  }

  Friday, June 15, 2007 9:11 AM
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  void main()

  {

   char far *farther,*farthest;

   

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

    

   }

  Friday, June 15, 2007 9:12 AM
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  main()

  {

   int i=400,j=300;

   printf("%d..%d");

  }

  Friday, June 15, 2007 9:12 AM
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  Linker Error : Undefined function 'Printf '

   

  If that is corrected then , the o/p will be i=0

   

  Friday, June 15, 2007 9:55 AM
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  senthil_vel_b10931 wrote:

  enum colors {BLACK,BLUE,GREEN} main() { printf("%d..%d..%d",BLACK,BLUE,GREEN); return(1); }

  
  
  0..1..2
  

  Friday, June 15, 2007 2:07 PM
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  senthil_vel_b10931 wrote:

  void main() { char far *farther,*farthest; printf("%d..%d",sizeof(farther),sizeof(farthest)); }

  
  4..2
  

  Friday, June 15, 2007 2:07 PM
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  senthil_vel_b10931 wrote:

  main() { int i=400,j=300; printf("%d..%d"); }

  
  
  
  300..400
  

  Friday, June 15, 2007 2:07 PM
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  senthil_vel_b10931 wrote:

  main() { clrscr(); } clrscr();

  
  
  
  No error. It will just clear screen.
  

  Friday, June 15, 2007 2:08 PM
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  senthil_vel_b10931 wrote:

  <!--[if !supportLists]-->1. <!--[endif]-->#include<stdio.h> main() { struct xx { int x=3; char name[]="hello"; }; struct xx *s; printf("%d",s->x); printf("%s",s->name); }

  
  
  Compiler Error
  

  Friday, June 15, 2007 2:11 PM
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  senthil_vel_b10931 wrote:

  <!--[if !supportLists]-->1. <!--[endif]-->main() { int i=5; printf("%d%d%d%d%d%d",i++,i--,++i,--i,i); }

  
  
  45545
  

  Friday, June 15, 2007 2:13 PM
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  senthil_vel_b10931 wrote:

  <!--[if !supportLists]-->1. <!--[endif]-->#define square(x) x*x main() { int i; i = 64/square(4); printf("%d",i); }

  
  
  64
  

  Friday, June 15, 2007 2:14 PM
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  senthil_vel_b10931 wrote:

  <!--[if !supportLists]-->1. <!--[endif]-->main() { char *p="hai friends",*p1; p1=p; while(*p!='\0') ++*p++; printf("%s %s",p,p1); }

  
  
  ibj!gsjfoet
  

  Friday, June 15, 2007 2:15 PM
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  senthil_vel_b10931 wrote:

  <!--[if !supportLists]-->1. <!--[endif]-->#include <stdio.h> #define a 10 main() { #define a 50 printf("%d",a); }

  
  
  50
  

  Friday, June 15, 2007 2:15 PM
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  senthil_vel_b10931 wrote:

  <!--[if !supportLists]-->1. <!--[endif]-->#define clrscr() 100 main() { clrscr(); printf("%d\n",clrscr()); }

  
  
  100
  

  Friday, June 15, 2007 2:15 PM
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  senthil_vel_b10931 wrote:

  <!--[if !supportLists]-->1. <!--[endif]-->main() { printf("%p",main); }

  
  
  address of main will be printed.
  

  Friday, June 15, 2007 2:15 PM
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  i think the GARBAGE VALUES will be printed.
  

  Saturday, June 16, 2007 1:31 PM
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  nice work man
  

  Friday, July 27, 2007 6:28 AM