can't get right output?

In my code everything works to a t but when you enter a 0 you get -1.jf? idk whats wrong.
any help would be appreaciated.

Code:

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#include "stdafx.h"
#include <conio.h>


int _tmain(int argc, _TCHAR* argv[])
{
        int x;
        int XCount;
        int sum;
        float average;

        sum = 0;
        XCount = 1;
        printf( "Enter Integer 1:" );
        scanf_s("%d", &x);

        while ( x != 0 ){
                        sum = sum + x;
                        XCount = XCount + 1;
                        printf("Enter Integer %d:", XCount);
                        scanf_s("%d", &x);
        }
                       


       
        average = ( float ) sum / (XCount - 1 );
        printf( "*****************\n" );
        printf( "You have entered %d non-zero numbers\n", XCount - 1);
        printf( "The sum of these integers is %d\n", sum );
        printf( "The average of these integers is %.2f\n", average );
        printf( "*****************" );

        getch();
        return 0;

}

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When you input 0, you have the calculation of:

Code:

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average = (float) 0 / 0;

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What do you expect the output of zero divided by zero to be?

0? haha idk. this is my first program.

0/0 is defined in math as equaling zero. But in programming dividing by zero is never OK. The actual answer you get is -1.#INF or -1.#NAN which means infinity, (or Not a Number) and that is usually correct but you can't use it. Print it unformatted (plain %f) and you will see.

Would this require me to post a separate print statement every time a value of 0 is given?

Quote:

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Originally Posted by jrg1424

Would this require me to post a separate print statement every time a value of 0 is given?

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That depends on what you mean. C can't do calculus like that so dividing by zero is an error you'd want to prevent. You could use printf to say something like "I can't divide by zero."

Quote:

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Originally Posted by whiteflags

That depends on what you mean. C can't do calculus like that so dividing by zero is an error you'd want to prevent. You could use printf to say something like "I can't divide by zero."

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yeah, i added an if statement using the XCount and it worked. thanks for the help

Code:

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#include "stdafx.h"
#include <conio.h>


int _tmain(int argc, _TCHAR* argv[])
{
        int x;
        int XCount;
        int sum;
        float average;

        sum = 0;
        XCount = 1;
        printf( "Enter Integer 1:" );
        scanf_s("%d", &x);

        while ( x != 0 ){
                        sum = sum + x;
                        XCount = XCount + 1;
                        printf("Enter Integer %d:", XCount);
                        scanf_s("%d", &x);
        }
                       


        if (XCount != 1 ){
        average = ( float ) sum / (XCount - 1 );
        printf( "*****************\n" );
        printf( "You have entered %d non-zero numbers\n", XCount - 1);
        printf( "The sum of these integers is %d\n", sum );
        printf( "The average of these integers is %.2f\n", average );
        printf( "*****************" );
}
      else{
        printf( "*****************\n" );
        printf( "You have entered %d non-zero numbers\n", XCount - 1);
        printf( "The sum of these integers is 0\n");
        printf( "The average of these integers is 0\n");
        printf( "*****************" );
}

        getch();
        return 0;

}

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