C static array pointer address

Question

We have a code.

#include <stdio.h>

int* aa(int s){
    static int ad[2] = {0};
    ad[0] = s;
    printf("aa() -> %p\n", &ad);
    return ad;
}

int main(void) {
    int *k = aa(3);
    printf("main() -> %p\n", &k);
    return 0;
}

Compile, run.. output

aa() -> 0x80497d0
main() -> 0xbfbe3e0c

or i misunderstood or this code have problems.

we are returning the same address for static array ad why on output they differ?

Barmar · Accepted Answer · 2015-04-14 16:52:15Z

4

In main you're printing the address of the k pointer variable, not the address of the array that it points to. If you do:

printf("main() => %p\n", k);

you'll get the same address as printed in aa.

answered Apr 14, 2015 at 16:52

Barmar

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2 Comments

Denis Balan Over a year ago

ok, but technically in low-end address of k isn't the address of ad? so they might be the same, or no?

Barmar Over a year ago

Yes, it is. When you return the array as a value, it's automatically converted to the address of the first element, just like when you pass it as an argument to printf.

Ekleog · Accepted Answer · 2015-04-14 16:51:53Z

2

You wrote printf "&k" which should just be "k". Hence you displayed the address of the variable k, and not of ad. :

#include <stdio.h>

int* aa(int s){
    static int ad[2] = {0};
    ad[0] = s;
    printf("aa() -> %p\n", &ad);
    return ad;
}

int main(void) {
    int *k = aa(3);
    printf("main() -> %p\n", k);
    return 0;
}

answered Apr 14, 2015 at 16:51

Ekleog

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Comments

Arun A S · Accepted Answer · 2015-04-14 16:52:46Z

1

&k returns the address of k and not the address it is pointing to. So, what you are getting is the address of k. You should change your printf() to

printf("main() -> %p\n", k);
                        ^
                        No need for &

answered Apr 14, 2015 at 16:52

Arun A S

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