pointer

Question

#include<stdio.h>
int main()
{
    int a = 10;
    void *ptr = &a;
    printf("%d", sizeof(ptr));
    return 0;
}
output =8

how ?if we know size of pointer is already machine dependent then in this case ptr is referring a address of int  then how output 8 ?

Comments

Sizeof is an operator which returns the amount of memory reserved for storing the value of a certain Data Type or a Variable. Since a pointer (to any data type) ultimately stores an address of some memory location, it reserves sufficiently many bytes that are required to address a location in memory.

In 64-bit machines, memory is addressed using 64 bits i.e., 8 Bytes. Now if I had to store an address of a location in memory at some other location in memory, I will need 8 Bytes of space. This is what the sizeof is returning. 

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In case you are wondering what difference does void data type make to the pointer, take this example

#include<stdio.h>
void main()
{
    int a = 10;
    void *ptr = &a;
    char *ptr2 = (char *)ptr + 1;
    int *ptr3 = (int *)ptr + 1;
    printf("%p %p %p\n", ptr,  ptr2, ptr3);
}   

Casting void pointer to an int or any other data structure doesn't change the storage capacity of a pointer. However, when the void pointer is cast to an int, it will fetch data from two subsequent memory location (going by convention, though this is machine dependent) when dereferencing the pointer.