how to compile the blue part?

This is a discussion on how to compile the blue part? within the C++ Programming forums, part of the General Programming Boards category; prob2(): By using a pointer to pointer **A and the function malloc() allocate the memory for the 4x4 matrix A[][]. ...

  1. #1
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    how to compile the blue part?

    prob2(): By using a pointer to pointer **A and the function malloc()
    allocate the memory for the 4x4 matrix A[][]. By using a pointer *b
    and the function malloc() allocate the memory for the 4-dimensional vector
    b[]. Read the components of A and b from the given input file matrix.dat.
    The last line of the input file contains the components of b. The rows of
    matrix A are the first four lines of the input file. Print the components
    of A[][] row by row, and the components of b[]. Using the function gauss()
    given in the lecture notes week9.txt, solve the system of linear algebraic
    equations A[][]*x[]=b[]. Print the components of the solution vector x[].
    Calculate and print the length of this vector. Free the allocated memory.
    ..


    the code that i constructed is as followed starting with the gauss
    Code:
    double **A, *b;
          int row;
            s=gauss(4,A,b);
            printf("x[]={%7.3f", s);
    
            for(row=0; row<4; row++) free(A[row]);
            printf("\nThe length of x[] is");
            free(A);
            free(b);
            fclose(input);
    }
            return;
    }
    
    
    double gauss(int n, double **A, double *b)
            {
            double one = 1.0, zero=0.0;
            double i, j;
            double s;
            int nm, row, col, Arow;
            nm = n-1;
            if(n==1)
            {
            b[0] /= A[0][0]; A[0][0] = one;
            return;
            }
            for(row=0; row<nm; row++)
            {
            i = A[row][row]; A[row][row] = one;
            for(col=row+1; col<n; col++) A[row][col] /=i;
            b[row] /=i;
            for(Arow=row+1; Arow<n; Arow++)
            {
            j=A[Arow][row]; A[Arow][row]=zero;
            for(col=row+1; col<n; col++) A[Arow][col] -= j * A[row][col];
            b[Arow] -= j * b[row];
            }
            }
            b[nm] /= A[nm][nm];
            for(row=nm-1; row>=0; row--)
            {
            for(col=row+1; col<n; col++) b[row] -=A[row][col]*b[col];
            s=b[row];
            }
            return s;
    Last edited by Salem; 03-10-2010 at 10:33 AM. Reason: Added [code][/code] tags - learn to use them yourself

  2. #2
    Programming Wraith GReaper's Avatar
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    Whata do you mean? What's the problem?
    Devoted my life to programming...

  3. #3
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    I tried to printf x[];
    but the result gave me nothing! is there something wrong with "double gauss(int n, double **A, double *b)?

  4. #4
    The larch
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    You never allocate any memory for A and b, only free it.

    Also the code is a bit fragmented, and you should use a [code] [/code] box.
    I might be wrong.

    Thank you, anon. You sure know how to recognize different types of trees from quite a long way away.
    Quoted more than 1000 times (I hope).

  5. #5
    and the hat of wrongness Salem's Avatar
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    Meh, added tags, but the indentation is still crap.
    SourceForge.net: Indentation - cpwiki
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
    I support http://www.ukip.org/ as the first necessary step to a free Europe.

  6. #6
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    i still dont understand...if you were to write it? how would you structure it? since it uses fscanf/sscanf...im confused

  7. #7
    and the hat of wrongness Salem's Avatar
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    Perhaps if the code was indented half-way decent,
    a) you might figure out the answer
    b) people might be more inclined to help you.

    Code:
    double gauss(int n, double **A, double *b)
    {
      double one = 1.0, zero=0.0;
      double i, j;
      double s;
      int nm, row, col, Arow;
    
      nm = n-1;
    
      if(n==1)
      {
        b[0] /= A[0][0]; 
        A[0][0] = one;
        return;
      }
    
      for(row=0; row<nm; row++)
      {
        i = A[row][row]; 
        A[row][row] = one;
        for(col=row+1; col<n; col++) 
          A[row][col] /=i;
        b[row] /=i;
    
        for(Arow=row+1; Arow<n; Arow++)
        {
          j=A[Arow][row]; 
          A[Arow][row]=zero;
          for(col=row+1; col<n; col++) 
            A[Arow][col] -= j * A[row][col];
          b[Arow] -= j * b[row];
        }
      }
    
      b[nm] /= A[nm][nm];
      for(row=nm-1; row>=0; row--)
      {
        for(col=row+1; col<n; col++) 
          b[row] -=A[row][col]*b[col];
        s=b[row];
      }
      return s;
    }
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
    If at first you don't succeed, try writing your phone number on the exam paper.
    I support http://www.ukip.org/ as the first necessary step to a free Europe.

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