Program will not run thorugh the loop

This is a discussion on Program will not run thorugh the loop within the C Programming forums, part of the General Programming Boards category; Hello The program asks the question one time then stops and creates a micorsoft error message. Could someone take a ...

  1. #1
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    Program will not run thorugh the loop

    Hello
    The program asks the question one time then stops and creates a micorsoft error message. Could someone take a look at it and see if I am missing something. It is suppose to allow you to enter up to 10 accouts w/name and amounts and break if -999 is entered.

    Code:
    
    #include <stdio.h>
    struct info
    {
     int   client_num;
     char  last_name[20];
     float balance;
    };
    
    main()
    {
    
     struct info clients[10];
     int    x;
    
     printf ("Enter account number, last name, and balance.\nEnter -999 to end input.\n\n");
    
     for (x = 0; x < 10; x++)
     {   
    
     printf ("?");
     scanf ("%i", clients[x].client_num);
      if (clients[x].client_num == -999)
     {
     break;
     }
    
     scanf ("%s", &clients[x].last_name);
     scanf ("%f", &clients[x].balance);
     fflush(stdin);
    
    printf ("\n\n");
    }
    
    printf ("ACCOUNT \tLAST NAME \tBALANCE\n");
    
    for (x = 0; x < 10; x++)
     {   
    
    printf ("%08i%20s%.102f\n", clients[x].client_num, clients[x].last_name, clients[x].balance);
     }
    getchar();
    }

  2. #2
    and the hat of wrongness Salem's Avatar
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    > scanf ("&#37;i", clients[x].client_num);
    Here you forgot the &

    > scanf ("%s", &clients[x].last_name);
    Here you used an & where one wasn't necessary.

    > fflush(stdin);
    See the FAQ, this is undefined (even if it appears to work for you).
    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.

  3. #3
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    Thank you, I knew it was something stupid. With the fflush(stdin); that is what the teacher requests which is why I used it.

  4. #4
    and the hat of wrongness Salem's Avatar
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    > With the fflush(stdin); that is what the teacher requests which is why I used it.
    Not a good sign.
    http://c-faq.com/stdio/stdinflush.html

    What other horrors are they going to teach you I wonder.
    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.

  5. #5
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    Can I bother you about one more thing??? It is now reading all of my info but if I enter the -999 it is breaking but not going to the next print statment. I thought it would just break in the current for statement and then move to the next one???

  6. #6
    and the hat of wrongness Salem's Avatar
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    Post your latest code.
    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.

  7. #7
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    Code:
    #include <stdio.h>
    struct info
    {
     int   client_num;
     char  last_name[20];
     float balance;
    };
    
    main()
    {
    
     struct info clients[10];
     int    x;
    
     printf ("Enter account number, last name, and balance.\nEnter -999 to end input.\n\n");
    
     for (x = 0; x < 10; x++)
     {   
    
     printf ("?");
     scanf ("&#37;i", &clients[x].client_num);
      if (clients[x].client_num == -999)
         break;
    
     scanf ("%s", clients[x].last_name);
     scanf ("%f", &clients[x].balance);
     fflush(stdin);
    }
    printf ("\n\n");
    
    
    printf ("ACCOUNT LAST NAME BALANCE\n");
    
    for (x = 0; x < 10; x++)
     {   
    
    printf ("%08i%20s%.102f\n", clients[x].client_num, clients[x].last_name, clients[x].balance);
     }
    getchar();
    }

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