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

Question

For example: sizeof(char*) returns 4. As does int*, long long*, everything that I've tried. Are there any exceptions to this?

Answer 1

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

Answer 2

From what I recall, it's based on the size of a memory address. So on a system with a 32-bit address scheme, sizeof will return 4, since that's 4 bytes.

Answer 3

The reason the size of your pointer is 4 bytes is because you are compiling for a 32-bit architecture. As FryGuy pointed out, on a 64-bit architecture you would see 8.

Answer 4

In general, sizeof(pretty much anything) will change when you compile on different platforms. On a 32 bit platform, pointers are always the same size. On other platforms (64 bit being the obvious example) this can change.

Answer 5

A pointer is just a container for an address. On a 32 bit machine, your address range is 32 bits, so a pointer will always be 4 bytes. On a 64 bit machine were you have an address range of 64 bits, a pointer will be 8 bytes.

Answer 6

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.

Answer 7

No, the size of a pointer may vary depending on the architecture. There are numerous exceptions.

Answer 8

In addition to the 16/32/64 bit differences even odder things can occur.

There have been machines where sizeof(int *) will be one value, probably 4 but where sizeof(char *) is larger. Machines that naturally address words instead of bytes have to "augment" character pointers to specify what portion of the word you really want in order to properly implement the C/C++ standard.

This is now very unusual as hardware designers have learned the value of byte addressability.

Answer 9

It will always be equal to sizeof size_t.

On 32-bit machines this tends to be 4. On 64-bit machines this tends to be 8.

Answer 10

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

Answer 11

In addition to what people have said about 64-bit (or whatever) systems, there are other kinds of pointer than pointer-to-object.

A pointer-to-member might be almost any size, depending how they're implemented by your compiler: they aren't necessarily even all the same size. Try a pointer-to-member of a POD class, and then a pointer-to-member inherited from one of the base classes of a class with multiple bases. What fun.

Answer 12

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.

Answer 13

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.