Character array and address of character array

Question

So if I have the code:

void main (void)
{
     char s[] = "Programming is hard";

    printf("%s", &s);
    printf("%s", s);
    return 0;

}

They both produce the same result. But &s should be the address of the character array right? so shouldnt' printf prints out the adress instead of what is stored in the address?

Thanks.

By definition (the value of an) address of an array is the same (value as) the address of the array's 1st element. — alk
– alk, Commented Jan 14, 2017 at 14:12
the '%s' prints what is at where the address points. to print out the actual address use : '%p' — user3629249
– user3629249, Commented Jan 15, 2017 at 15:14

Bathsheba · Accepted Answer · 2017-01-14 18:09:56Z

2

The behaviour of the first printf is undefined due to a mismatched formatter.

In many respects, your compiler is being kind to you.

Don't do that!

edited Jan 14, 2017 at 18:09

answered Jan 14, 2017 at 14:10

Bathsheba

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Comments

XXXBold · Accepted Answer · 2017-01-14 14:26:51Z

-1

If you want to print out an address of a variable, use %x or %p as formatter.

answered Jan 14, 2017 at 14:26

XXXBold

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2 Comments

alk Over a year ago

x is not correct unless the pointer value is casted to best uintptr_t.

alk Over a year ago

Also when using p the pointer value needs to be casted to void*.

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