C error, simple scanf functions with integers giving 0 as result

Question

I'm experimenting and training some C and I found out a very strange "error" that I can't understand.

My code is:

#include <stdio.h>

int main() {

    int num1, num2;
    int num3 = num1 * num2;

    printf("Enter a number: ");
    scanf("%d", &num1);
    printf("Enter one more number: ");
    scanf("%d", &num2);
    printf("The result of %d * %d is %d\n", num1, num2, num3);
}

I want it to assume that the variable num3 is the product between num1 and num2.

I got this output:

Enter a number: 8
Enter one more number: 7
The result of 8 x 7 is 0

Numbers 8 and 7 are random numbers I typed in.

Why is it 0? It is supposed to be 56.

Answer 1

You are computing num3, even before num1 and num2 have been read in from the command line. Assuming, num1 and num2 were initialized to zero num3 would be zero at the time of computation.

Answer 2

Filipe, you have got to put the horse before the cart. Variables must be initialized with a value before you multiply them together. (in fact, attempting to access a value in an uninitialized variable invokes Undefined Behavior (that's bad....)

Just what do you expect to happen in the following code?

int num1, num2;
int num3 = num1 * num2;

What values are contained in num1, num2? What do you think will get put in num3? In instances like this it is really a good idea to see How to debug small programs and talk to the duck... Really, it helps :)

Next there is unvalidated user input, e.g.

scanf ("%d", &num1);

What happens if the user enters "one" (or a cat steps on the keyboard)? Always, always validate user input. scanf returns the number of successful conversions that take place. In the case of "%d" -- that's 1 if the command successfully places an integer value in num1. A simple test can save a lot of grief, e.g.

    printf ("Enter a number: ");
    if (scanf ("%d", &num1) != 1) { /* validate ALL user input */
        fprintf (stderr, "error: invalid input or user canceled.\n");
        return 1;
    }

By checking that the returns of scanf is 1, you protect against both matching and input failures (see: man scanf).

Only after you have successfully placed values in num1 and num2 can you validly multiply num1 and num2 together, e.g.

    num3 = num1 * num2; /* num1 & num2 must have values first */

Putting all the pieces together, the horse before the cart, and noting that main() is type int and therefore should return a value ^[1], you could do something similar to the following:

#include <stdio.h>

int main (void) {

    int num1, num2, num3;

    printf ("Enter a number: ");
    if (scanf ("%d", &num1) != 1) { /* validate ALL user input */
        fprintf (stderr, "error: invalid input or user canceled.\n");
        return 1;
    }

    printf ("Enter one more number: ");
    if (scanf ("%d", &num2) != 1) { /* Ditto */
        fprintf (stderr, "error: invalid input or user canceled.\n");
        return 1;
    }

    num3 = num1 * num2; /* num1 & num2 must have values first */

    printf ("The result of %d * %d is %d\n", num1, num2, num3);

    return 0;   /* main() is type int and returns a value. */
}

Example Use/Output

$ ./bin/multiply
Enter a number: 8
Enter one more number: 7
The result of 8 * 7 is 56

Look things over and let me know if you have further questions.

footnote[1] See: C11 Standard §5.1.2.2.1 Program startup (draft n1570). See also: See What should main() return in C and C++?

Answer 3

You could also add num3 to first init then after prompting the user to enter numbers have the line num3=num1*num2;