Thread: Why is the output 9?

You probably should have written:

Code:

#include<stdio.h>

void f(int *w)
{
    *w = *w + sizeof(w);
}

int main(void)
{
    int a = 5;
    f(&a);
    printf("%d\n", a);
}

Notice that:

• #include<string.h> was removed as none of the functions declared within it were used.
• The code for function f has been formatted properly.
• main has been declared as int main(void), not just as main(). You should declare the return type, and while it is optional in a function definition, you should declare the parameter list as void if the function takes no arguments.
• f(a) has been changed to f(&a). f(a) is wrong since a is an int, but f expect a pointer to int as the argument.
• I kept to the leaving out of the return statement from main as it is a special case since C99, but if you are compiling with respect to pre-C99, you should have a return 0; at the end of main.

Originally Posted by SauceGod

Can someone explain to me why the output of this program is 9?

Since &a was passed to f, w points to a, so *w is a, i.e., it has a value of 5. sizeof(w) presumably is 4 for you, so the result is 9. Note that the size of a pointer is not necessarily 4 bytes, so you could end up with a different valid result, e.g., if sizeof(w) == 8, the result would be 13.

It seems "why is the result XYZ?" is becoming your theme, OP. Wouldn't it make sense to experiment and try to find out why these results are happening?

Code:

#include <stdio.h>
void f(int *w)
{
    *w = *w + sizeof(w);
}
int main(void)
{
    int a = 5;
    printf("value of a=%d\n", a);
    printf("value of sizeof(&a)=%d\n", (int) sizeof(&a));
    f(&a);
    printf("result of f(&a)=%d\n", a);
    return 0;
}

Just an idea. Sometimes a little inspection goes a long way.