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MFC appliation RRS feed

  • Question

  • Hi all,

    i am facing a strange problem with an MFC appliation.

    The problem is the application is giving different ouputs for the same input on different runs.

    what might be the cause for this unpredictabiligy?

    Appliations takes in 2^n bits input and computes walsh transform.

    when i run the application first time i have one output

    when i run the application with the same input i had different output

    Could u tell some possible reasons for this unpredictability.

    The application reads two inputs from the user n and a file consists of only one line of bits.

    The code has given below:

    The functin onButtonClicked() contains the following

    {

    UpdateData(TRUE);

      int *W_f,x, *Bfun,i;

    x=pow(2,n);

    fstream fpout,fpin;

    fpin.open("truthtable.txt",ios::in);

    fpout.open("output.txt",ios::out);

      //Allocate memory       

      W_f =new  int [x];

      Bfun = new int [x];

    //Read truth table of boolean funciton f from the file

      string fs;

      getline(fpin,fs);                          
      for(  i =0;i<x;i++)
         {
                    if(fs[i]=='1')
                        f[i]=1;
                    else
                        f[i]=0;
       }    

    //Compute fast walsh transform

      FastWalshTranform(Bfun, n, W_f);

      for(i=0;i<x;i++)

          fpout<<W_f[i]<<"\n";

    }

    void FastWalshTranform(int *Bfun, int n, int &W_f)
    {

    *int UD,A,B;
    int i,j,x,x_half,k,l,b,w,p;
    x=(int)pow(2,n);
    x_half=x/2;
    UD=new int [x];
    //CopyTruth table of f into UD
    for(i=0;i<x;i++)
    {
        if(Bfun[i]==1)
            UD[i]=1;                        
        else
            UD[i]=0;
    }                
        //FastWalshTransform
        for(k=0;k<n;k++)
        {
            l=(int)pow(2,k);
            b=x/l;
            p=b/2;        
            A=new int [l];
            B=new int [l];
            for(i=0,j=0; i<p; i++,j=j+2*l)
            {
                for(w=0;w<l;w++)
                {
                    A[w]=UD[j+w];
                    B[w]=UD[j+l+w];
                }                    
                for(w=0;w<l;w++)
                {
                    UD[j+w]=A[w]+B[w];
                    UD[j+l+w]=A[w]-B[w];
                }                    
                            
            }//end of for(i=0;j=0; i<p; i++,j=j+2*l)
            delete A[];
            delete B[];
        }//end of for(k=0;k<n;k++)
          //COmpute Walsh Transform
        W_f[0]=x-2*UD[0];
        for(w=1;w<x;w++)
                W_f[w]=x - 2*( UD[w]+x_half );                                        
        //Free space for UD
        delete UD [];
    }

    with regards

    Ganesh Y



    Monday, May 28, 2012 5:40 AM

Answers