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

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 2

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 3

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.

Answer 4

In Win64 (Cygwin GCC 5.4), let's see the below example:

First, test the following struct:

struct list_node{
    int a;
    list_node* prev;
    list_node* next;
};

struct test_struc{
    char a, b;
};

The test code is below:

std::cout<<"sizeof(int):            "<<sizeof(int)<<std::endl;
std::cout<<"sizeof(int*):           "<<sizeof(int*)<<std::endl;
std::cout<<std::endl;

std::cout<<"sizeof(double):         "<<sizeof(double)<<std::endl;
std::cout<<"sizeof(double*):        "<<sizeof(double*)<<std::endl;
std::cout<<std::endl;

std::cout<<"sizeof(list_node):      "<<sizeof(list_node)<<std::endl;
std::cout<<"sizeof(list_node*):     "<<sizeof(list_node*)<<std::endl;
std::cout<<std::endl;

std::cout<<"sizeof(test_struc):     "<<sizeof(test_struc)<<std::endl;
std::cout<<"sizeof(test_struc*):    "<<sizeof(test_struc*)<<std::endl;    

The output is below:

sizeof(int):            4
sizeof(int*):           8

sizeof(double):         8
sizeof(double*):        8

sizeof(list_node):      24
sizeof(list_node*):     8

sizeof(test_struc):     2
sizeof(test_struc*):    8

You can see that in 64-bit, sizeof(pointer) is 8.

Answer 5

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 6

Size of pointer and int is 2 bytes in Turbo C compiler on windows 32 bit machine.

So size of pointer is compiler specific. But generally most of the compilers are implemented to support 4 byte pointer variable in 32 bit and 8 byte pointer variable in 64 bit machine).

So size of pointer is not same in all machines.