获取变量的十六进制地址为uintptr_t

Question

int y = 1;
int *x = &y;  
printf("%p\n",x); // instead of printing, get this into variable of type unintptr_t

I would like to get the address x into variable of type uintptr_t

Is there a way to do that in C?

Answer 1

It isn't particularly difficult...

uintptr_t z = (uintptr_t)x;

Notice that the result of this cast is implementation-defined; the only guarantee you have is that if you cast z back to an int * you'll get the original pointer back.

By the way, there's no such a thing as a "hexadecimal address"; addresses are addresses, they can be seen as numbers in whatever base you like most, showing them in hexadecimal base is just a convention (which has some advantages).

Answer 2

您可以通过将x的地址显式转换为uintptr_t来存储值。

uintptr_t z = (uintptr_t)&x;