将32位指针转换为64位整数

Question

What's the best way to do this?

Say I have:

 void* p = 0xc8f68000;
 unsigned long long v = (unsigned long long)p;

I got v = 0xffffffffc8f68000 instead of 0x00000000c8f68000. Do I have to use shift?

Answer 1

Whenever you deal with pointer to integer conversions, it's best to use uintptr_t . It is the portable way to avoid size and sign extension issues when casting pointers.

#include <stdint.h>

void *p = 0xc8f68000;
uint64_t v = (uintptr_t)p;

Answer 2

What seems to be happening is that 0xc8f68000 is being treated as a signed 32 bit integer, and then being converted to unsigned. You should get the 'correct' value with:

void* p = 0xc8f68000;
unsigned long long v = (unsigned long)p;

or probably better:

void* p = 0xc8f68000;
unsigned long long v = (size_t)p;

since the latter isn't dependent on p actually being 32 bits.

I question why you want to do this, however.