CODEWITHSUNDEEP
c
Avinash
Avinash

Reputation: 13257

Passing Type to C macros

I am trying to pass a type in C macro, But I am getting error.

#include <stdio.h>)

    int size = 0;
    #define SIZEOF(TYPE) \
        TYPE _array_[2]; \
        size = (char*)(&_array_[1]) - (char*)(&_array_[0]);

    int main(int argc, char** argv, char** envp) {
        int array[2];
        int x = SIZEOF(int);
        printf("Size Of Integer = %d", SIZEOF(int));
        return 0;
    }

Upvotes: 0

Views: 203

Answers (2)

M.M
M.M

Reputation: 141554

Macro expansion is a text replacement. Your code expands to:

int x = int _array_[2]; size = (char*)(&_array_[1]) - (char*)(&_array_[0]);

The initial part int x = int is a syntax error.

You could make your code work without changing the macro by writing:

SIZEOF(int);
printf("Size Of Integer = %d", size);

An improvement would be to pass the name of the variable as a parameter to the macro, instead of using global variables.

Upvotes: 3

juhist
juhist

Reputation: 4314

Try this:

#define SIZEOF(TYPE) \
    ({ \
    TYPE _array_[2]; \
    (char*)(&_array_[1]) - (char*)(&_array_[0]); \
    })

...although it uses gcc specific support for statements inside expressions, so I would recommend using the standard sizeof operator instead.

How to actually use this:

printf("Size Of Integer = %d", (int)SIZEOF(int));

i.e. it works as a pretty good replacement for sizeof on compilers that actually allow the dirty trick of statements inside expressions.

Upvotes: 1

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