Why is sizeof(int) different than sizeof(int*)?

Question

I am wondering why in the following program sizeof(int) returns a different value than sizeof(int*).

Here is the small program:

int main(){
    std::cout<<sizeof(int)<<endl;
    std::cout<<sizeof(int*)<<endl;
    return 0;
}

And here is the output:

4
8

Till now I remember the size of a integer pointer is 4byte(gcc compiler). How can I check the correct size of a pointer? Is it computer dependent?

I am running ubuntu 12.04

# lsb_release -a

Distributor ID: Ubuntu 
Description: Ubuntu 12.04 LTS 
Release:    12.04 
Codename:   precise

Is the size of pointer is not constant(standard size) 8 bytes.

Answer 1

The size of an int and an int* are completely compiler and hardware dependent. If you're seeing eight bytes used in an int*, you likely have 64-bit hardware, which translates into eight bytes per pointer.

Hope this helps!

Answer 2

sizeof(char) == 1

There are no other guarantees(*).

In practice, pointers will be size 2 on a 16-bit system, 4 on a 32-bit system, and 8 on a 64-bit system.

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(*) See the comment of James Kanze.

Answer 3

The size of a pointer is system, compiler, and architecture-dependent. On 32-bit systems it will typically be 32 bits while on 64-bit systems they will typically be 64 bits.

If you're trying to store a pointer into an integer for later restoration into the pointer again you can use the type intptr_t which is an integral type big enough to hold (I believe) normal (non-function) pointer types.