Thread: simple program giving me issues:

  1. #1
    Registered User
    Join Date
    Jul 2013

    simple program giving me issues:

    All I want to do is divide the smaller number into the larger one. I has no issues test and getting the right answer. Now I wanted to pass the remainder and this is where the issues came.

    After I can effectively get the remainder passed to (int sum2) I want to divide the smaller remainder into the larger one until the remainder is 0. I'll use a "while" statement for this.

    Right now if I divide 9 / 3 = 0 r 3
    3 / 9 = 0 r 3

    5 / 5 = 32767 R 1606416048

    This is a very simple program and I going in circles here. Can anyone help.

    int remainder (int, int);           // function protoype //
    #include <stdio.h>
    int remainder(int x, int y)         // function header //
        int sum1, sum2;
        if (x > y) {                    // the lesser divided into the highier // 
            sum1 = y / x;
            sum2 = y%x;                 // the remainder is passed to sum2 //
        if (y > x) {
            sum1 = x / y;
            sum2 = x%y;
        return printf("The answer is %d and the remainder is %d", sum1, sum2);
    int main(int argc, const char * argv[])
        int x,y;
        printf("Enter two numbers\n");
        scanf("%d%d",&x, &y);
        remainder(x, y);        // function called //
        return 0;
    Last edited by jocdrew21; 07-13-2013 at 01:38 AM.

  2. #2
    Hurry Slowly vart's Avatar
    Join Date
    Oct 2006
    Rishon LeZion, Israel
    Well - what do you think your program should print when y==x?
    All problems in computer science can be solved by another level of indirection,
    except for the problem of too many layers of indirection.
    David J. Wheeler

  3. #3
    Stoned Witch Barney McGrew's Avatar
    Join Date
    Oct 2012
    Three issues.

    If x == y, sum1 and sum2 will be referenced without having a value stored in them, which seems to be the problem you're most concerned with. I'm sure you can fix that.

    const char ** isn't compatible with char **, so your definition for main may not be allowed by your C implementation.

    If scanf returns a value that isn't 2, x or y will be referenced without having a value stored in them.

    EDIT: Also, you don't print a terminating new line to stdout.
    Last edited by Barney McGrew; 07-13-2013 at 02:30 AM.

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