从不同大小的整数转换为指针（将int作为const void *参数传递）

Question

I have to maintain a chunk of source code on Linux with gcc which has the following function prototype:

int foo(const void*, ...)

I have to pass an integer into this function. I know that this is a bad idea but at this stage, I'm not allowed to modify the source code.

ie I have to call foo(n) ( n is defined as int n )

On a 64-bit machine, I get an error:

cast to pointer from integer of different size

As I understand, this is because the size of an integer on a 32-bit machine is four bytes while the size of a pointer to a void is eight bytes.

I resolve the compilation error by first casting the integer with size 4 bytes to a long with size 8 bytes. So I did foo((long)n)

I wonder if this is an acceptable way to deal with this issue or are there other suggestions?

Answer 1

You could use intptr_t , but it is optional and may not be available in stdint.h :

intptr_t n = 1;
foo((void*)n);