What's the best portable way to represent pointer as string in C++?

Question

I need to represent pointers as strings to the user. Sometimes the values might be saved to a file and transferred to a computer with different architecture (32 vs 64 bit is the main issue currently) and loaded from text file to be compared - I'm only going to compare loaded values with each other, but I'd still prefer to compare numbers than strings.

I'm currently using:

SomeClass* p;
...
printf("%ld", (uintptr_t)p);

but I wonder if this is portable (Windows and Linux are only important at this stage though), and whether this would break once 128-bit systems show up?

Edit: unless I decide to use uint64_t, and decide that 64bit is the rooftop, this cannot be done because some 64bit pointer might be outside 32bit integer range. So, I decided that it would be safer to compare strings even if it's slower.

Answer 1

For pointers, always use %p---it's a format specifier specially designed for printing pointers in the right format. :-)

Answer 2

printf has a %p formatter which I believe is standardised:

printf( "%p", p );

but as you are using C++, ostreams already overload pointer output:

#include <iostream>
using namespace std;;

class A {};

int main() {
    A a;
    cout << &a << endl;
}

produces:

0x22ff6f

Answer 3

I'd do this:

std::cout << p;

If you have your heart set on cstdio:

printf("%p", p);

Answer 4

Are you looking for the %p formatting string?

printf("%p", p);

This'll give the hex-encoded address the pointer points to (and, I think, formats NULLs for you as well).

Answer 5

I would have done:

cout << p << endl;

Strings (of the same encoding) are portable. If you want to compare them as pointers, you have to parse the string back into an unsigned long first.

As for 128-bit systems, they're a ways off. WIth 64 bits, you can directly address 16.8 million terabytes of RAM.

Answer 6

It can also be done as

printf("%#x", p);

This would ensure similar format (0x...) across platforms.