9 Digit Addresses in Hexadecimal System in MacOS

Question

I am practicing on C programming language. I had Win 10 systems and I was practicing on it and I always get addresses which are multiples of 2. Know, I am practicing on Macbook Pro M3 and I get 9 digit addresses. So, nine is odd.

Also, here is the code and result:

#include <stdio.h>

int main(void)
{
    int m = 10, n, o, *z;

    z = &m;

    printf("z stores the address of m = %p\n", z);
    printf("*z stores the value of m = %d\n", *z);
    printf("&m is the address of m = %p\n", &m);
    printf("&n stores the address of n = %p\n", &n);
    printf("&o stores the address of o = %p\n", &o);
    printf("&z stores the address of z = %p\n", &z);

    return 0;
}

z stores the address of m = 0x16fa13228
*z stores the value of m = 10
&m is the address of m = 0x16fa13228
&n stores the address of n = 0x16fa13224
&o stores the address of o = 0x16fa13220
&z stores the address of z = 0x16fa13218

I researched about it however I couldn't get good answers. Here what I found:

why the value of address in c s always even?

Strange address of a C variable in Mac OS X

I was expecting to see addresses in even digits.

Answer 1

From the C23 draft:

7.23.6.1 The fprintf function

8. The conversion specifiers and their meanings are:

• p The argument shall be a pointer to void or a pointer to a character type. The value of the pointer is converted to a sequence of printing characters, in an implementation-defined manner.

That means that different implementations may produce different output, even if the actual pointer values would happen to be the same.

I was expecting to see addresses in even digits.

I'm assuming this means that you expected to see an even number of digits. Typically, no more than the significant non-zero digits will be printed so if you get 9 digits on your Mac, that's just an effect of that. A 32 bit address like 0x00000FFF will often be printed as just 0xFFF.

---

If you want the same format on all platforms and implementations, you could cast the pointer to uintptr_t and print that instead.

Example:

#include <inttypes.h> // PRIxPTR
#include <stdint.h>   // uintptr_t
//...
printf("&n stores the address of n = 0x%016" PRIxPTR "\n", (uintptr_t)&n);

Possible output:

&n stores the address of n = 0x000000016fa13224

Answer 2

I had Win 10 systems and I was practicing on it and I always get addresses which are multiples of 2

Know, I am practicing on Macbook Pro M3 and I get 9 digit addresses.

It does not depend on a number of digits only the the value of that number.

Example: 0x16fa13228 is 6167802408 in decimal. This number is dividable by 2.