Is the sizeof(some pointer) always equal to four?
For example: sizeof(char*) returns 4. As does int* , long long* , everything that I've tried. Are there any exceptions to this?
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( "%dn", 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
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