Size of pointer of integer type vs Size of int*

Question

I started reading Pointers and while tinkering with them. I stumbled upon this :

#include<stdio.h>
int main()
{
    int *p,a;
    a=sizeof(*p);
    printf("%d",a);
}

It outputs : 4

Then in the place of sizeof(*p) I replaced it with sizeof(int*) Now it outputs 8 .

P is a pointer of integer type and int* is also the same thing ( Is my assumption correct? ). Then why it is printing two different values. I am doing this on a 64bit gcc compiler.

Answer 1

Every beginner always gets confused with pointer declaration versus de-referencing the pointer, because the syntax looks the same.

• int *p; means "declare a pointer to int". You can also write it as int* p; (identical meaning, personal preference).
• *p, when used anywhere else but in the declaration, means "take the contents of what p points at".

Thus sizeof(*p) means "give me the size of the contents that p points at", but sizeof(int*) means "give me the size of the pointer type itself". On your machine, int is apparently 4 bytes but pointers are 8 bytes (typical 64 bit machine).

Answer 2

*p and int* are not the same things! First one is a dereferenced pointer (i.e. int which is 4 bytes wide) and the second one is a pointer (8 bytes wide in your case since it's a 64 bit machine).