Is the sizeof(some pointer) always equal to four?

Question

The guarantee you get is that sizeof(char) == 1. There are no other guarantees, including no guarantee that sizeof(int *) == sizeof(double *).

In practice, pointers will be size 2 on a 16-bit system (if you can find one), 4 on a 32-bit system, and 8 on a 64-bit system, but there's nothing to be gained in relying on a given size.

Even on a plain x86 32 bit platform, you can get a variety of pointer sizes, try this out for an example:

struct A {};

struct B : virtual public A {};

struct C {};

struct D : public A, public C {};

int main()
{
    cout << "A:" << sizeof(void (A::*)()) << endl;
    cout << "B:" << sizeof(void (B::*)()) << endl;
    cout << "D:" << sizeof(void (D::*)()) << endl;
}

Under Visual C++ 2008, I get 4, 12 and 8 for the sizes of the pointers-to-member-function.

Raymond Chen talked about this here.

Just another exception to the already posted list. On 32-bit platforms, pointers can take 6, not 4, bytes:

#include <stdio.h>
#include <stdlib.h>

int main() {
    char far* ptr; // note that this is a far pointer
    printf( "%d\n", sizeof( ptr));
    return EXIT_SUCCESS;
}

If you compile this program with Open Watcom and run it, you'll get 6, because far pointers that it supports consist of 32-bit offset and 16-bit segment values

if you are compiling for a 64-bit machine, then it may be 8.

Technically speaking, the C standard only guarantees that sizeof(char) == 1, and the rest is up to the implementation. But on modern x86 architectures (e.g. Intel/AMD chips) it's fairly predictable.

You've probably heard processors described as being 16-bit, 32-bit, 64-bit, etc. This usually means that the processor uses N-bits for integers. Since pointers store memory addresses, and memory addresses are integers, this effectively tells you how many bits are going to be used for pointers. sizeof is usually measured in bytes, so code compiled for 32-bit processors will report the size of pointers to be 4 (32 bits / 8 bits per byte), and code for 64-bit processors will report the size of pointers to be 8 (64 bits / 8 bits per byte). This is where the limitation of 4GB of RAM for 32-bit processors comes from -- if each memory address corresponds to a byte, to address more memory you need integers larger than 32-bits.