implementation of sizeof operator in C

Question

I tried to implement what is given in this question.sizeof implementation

#include <stdio.h>
#include <stdint.h>

#define my_sizeof(type) ((char*)(&type + 1)-(char*)(&type))

int main()
{
    printf("Size of int   %d \n",my_sizeof(int));

    return 0;
}

However when I compile I get the following error.

test.c:10:44: error: expected expression before ‘int’
     printf("Size of int   %d \n",my_sizeof(int));
                                            ^
test.c:5:35: note: in definition of macro ‘my_sizeof’
 #define my_sizeof(type) ((char*)(&type + 1)-(char*)(&type))
                                   ^
test.c:10:44: error: expected expression before ‘int’
     printf("Size of int   %d \n",my_sizeof(int));
                                            ^
test.c:5:54: note: in definition of macro ‘my_sizeof’
 #define my_sizeof(type) ((char*)(&type + 1)-(char*)(&type))
                                                      ^

That expands to `((char*)(&int + 1)-(char*)(&int))` What is the address of int? Beats me... — John3136, Sep 15 '15 at 01:51

score 3 · Accepted Answer · answered Sep 15 '15 at 01:51

((char*)(&int + 1)-(char*)(&int))

Your macro is trying to take the address of a type. You could either make the macro a lot longer by including a whole block in it with a local variable (but then the macro wouldn't work how you want) or just only use the macro on variables and not types.

rcgldr · Answer 2 · 2015-09-15T02:15:46.673

2

This works for types, but not for variables:

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

edited Sep 15 '15 at 02:15

answered Sep 15 '15 at 02:07

rcgldr

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implementation of sizeof operator in C

2 Answers2