Question

#include <stdafx.h>
#include <stdio.h>
#include <conio.h>
#include<stdlib.h>
#define My_Sizeof(type) ((char*)((type*)0 +1) - (char*)(type*)0)
void main()
{
    char a='1';
    int b=My_Sizeof(int);
    printf("size is %d",b);
    _getch();
}

// can anybody help me to understand wt the macro does to calculate sizeof char datatype ?

Answer 1

Break it down:

#define My_Sizeof(type) ((char*)((type*)0 +1) - (char*)(type*)0)

(char*)(type*)0 is zero

(type*)0 +1 does pointer arithmetic using pointers of type (type *), so (type *)0 + 1 will be a pointer of offset exactly 0 + 1 * sizeof(type) = sizeof(type) bytes

When the difference is taken as (type *) the difference is 1. When the difference is taken with both types (char *), the difference is sizeof(T) - 0 = sizeof(T)