i need a debugger badly...

This is a discussion on i need a debugger badly... within the C Programming forums, part of the General Programming Boards category; Code: #include <stdio.h> #include <math.h> main() { double first_a, first_b, second_a, second_b, ans1, ans2, ans3, ans4, ans5, ans6; printf("Test Complex ...

  1. #1
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    i need a debugger badly...

    Code:
    #include <stdio.h>
    #include <math.h>
    
    main()
    {
      double first_a, first_b, second_a, second_b, ans1, ans2, ans3, ans4, ans5, ans6;
      printf("Test Complex Number Arithmetic");
      printf("\n");
      printf("Enter the first complex number in the form - real, imaginary - --> (");
      scanf(" ( %lf , %lf )", &first_a, &first_b );
      printf("\n");
      printf("Enter the second complex number in the form - real, imaginary - --> (");
      scanf(" ( %lf , %lf )", &second_a, &second_b );
      printf("\n");
      ans1 = (first_a + second_a, first_b + second_b);
      ans2 = (first_a - second_a, first_b - second_b);
      ans3 = (first_a * second_a, first_b * second_b);
      ans4 = (first_a / second_a, first_b / second_b);
      ans5 = (sqrt(first_a), sqrt(first_b));
      ans6 = (sqrt(second_a), sqrt(second_b));
      printf("Results:  (%lf) + (%lf) = (%lf)  (%lf) + (%lf) = (%lf)  (%lf) + (%lf) = (%lf)  abs(%lf) + (%lf) = (%lf)   abs(%lf) + (%i) = (%i)", first_a, first_b,
    second_a, second_b, ans1, first_a, first_b, second_a, second_b, ans2, first_a, first_b, second_a, second_b, ans2, first_a, first_b, second_a, second_b, ans3, first_a,$
    
    return 0;
    }
    Last edited by newcstudent; 10-12-2006 at 01:16 AM.

  2. #2
    Registered User whiteflags's Avatar
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    Please explain what is wrong if you can. Any information that you have, like a sample run of your program, your compiler, your OS, and your expectations are valuable.

  3. #3
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    Code:
    #include <stdio.h>
    #include <math.h>
    
    main()
    {
      double a1, b1, a2, b2, a_ans1, b_ans1, a_ans2, b_ans2, a_ans3, b_ans3, a_ans4, b_ans4, a_ans5, b_ans5, a_ans6, b_ans6;
      printf("Test Complex Number Arithmetic");
      printf("\n");
      printf("Enter the first complex number in the form - real, imaginary - -->");
      scanf(" ( %lf , %lf )", &a1, &b1 );
      printf("\n");
      printf("Enter the second complex number in the form - real, imaginary - -->");
      scanf(" ( %lf , %lf )", &a2, &b2 );
      printf("\n");
      a_ans1 = (a1 + a2);
      b_ans1 = (b1 + b2);
      a_ans2 = (a1 - a2);
      b_ans2 = (b1 - b2);
      a_ans3 = (a1 * a2);
      b_ans3 = (b1 * b2);
      a_ans4 = (a1 / a2);
      b_ans4 = (b1 / b2);
      a_ans5 = sqrt (a1);
      b_ans5 = sqrt (b1);
    
      printf("Results:\n"
             "(%lf + %lf) (%lf + %lf) = (%lf), (%lf)\n"
             "(%lf + %lf) (%lf + %lf) = (%lf), (%lf)\n"
             "(%lf + %lf) (%lf + %lf) = (%lf), (%lf)\n"
             "(%lf + %lf) (%lf + %lf) = (%lf), (%lf)\n"
             "(%lf + %lf) (%lf + %lf) = (%lf), (%lf)\n",
             a1, b1, a2, b2, a_ans1, b_ans1
             a1, b1, a2, b2, a_ans2, b_ans2
             a1, b1, a2, b2, a_ans3, b_ans3
             a1, b1, a2, b2, a_ans4, b_ans4
             a1, b1, a2, b2, a_ans5, b_ans5);
    
    return 0;
    this is my updated code... i need it to read something like the below...
    but i get a compiler error

    assignment3.c: In function `main':
    assignment3.c:33: error: syntax error before "a1"




    Test Complex Number Arithmetic

    Enter the first complex number in the form ( real, imaginary ) --> (1.0, 1.0)

    Enter the second complex number in the form ( real, imaginary ) --> (2.1, 3.4)

    Results:

    (1.000000, 1.000000) + (2.100000, 3.400000) = ( 3.100000e+000, 4.400000e+000)
    (1.000000, 1.000000) - (2.100000, 3.400000) = (-1.100000e+000, -2.400000e+000)
    (1.000000, 1.000000) * (2.100000, 3.400000) = (-1.300000e+000, 5.500000e+000)
    (1.000000, 1.000000) / (2.100000, 3.400000) = ( 3.443957e-001, -8.140263e-002)
    abs(1.000000, 1.000000) = (1.414214)
    abs(2.100000, 3.400000) = (3.996248)
    Last edited by newcstudent; 10-12-2006 at 03:36 AM.

  4. #4
    Registered User SKeane's Avatar
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    Code:
     printf("Results:\n"
             "(%lf + %lf) + (%lf + %lf) = (%lf, %lf)\n"
             "(%lf + %lf) - (%lf + %lf) = (%lf, %lf)\n"
             "(%lf + %lf) * (%lf + %lf) = (%lf, %lf)\n"
             "(%lf + %lf) / (%lf + %lf) = (%lf,  %lf)\n"
             "(%lf + %lf) (%lf + %lf) = (%lf, %lf)\n",
             a1, b1, a2, b2, a_ans1, b_ans1,
             a1, b1, a2, b2, a_ans2, b_ans2,
             a1, b1, a2, b2, a_ans3, b_ans3,
             a1, b1, a2, b2, a_ans4, b_ans4,
             a1, b1, a2, b2, a_ans5, b_ans5);
    Note the ',' on the end of each variable list staring a1, ...

  5. #5
    and the hat of wrongness Salem's Avatar
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    a1, b1, a2, b2, a_ans1, b_ans1
    a1, b1, a2, b2, a_ans2, b_ans2
    a1, b1, a2, b2, a_ans3, b_ans3
    a1, b1, a2, b2, a_ans4, b_ans4
    a1, b1, a2, b2, a_ans5, b_ans5);

    Add more commas
    Each parameter needs a comma, even those at the end of a line

    Edit: beaten
    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.
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  6. #6
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    thanks. it helped alot.

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