Issue

I have a macro that uses GCC's typeof to create a variable of the same type of a macro argument. The problem is: if that argument has const type, the variable created inside the macro is const and I can't use it. For instance:

#include <stdio.h>

#define DECR(x) ({typeof(x) y; y = x; y--; y;})

int main(void)
{
    const int v = 5;
    printf("%d\n", DECR(v));
    return 0;
}

Compilation gives:

$ cc    -c -o t.o t.c
t.c: In function 'main':
t.c:9:2: error: assignment of read-only variable 'y'
t.c:9:2: error: decrement of read-only variable 'y'
make: *** [t.o] Error 1

Is there a way to copy the typeof a value and un-const it?

Solution

If you don't mind the possible arithmetic promotion you can do this:

#define DECR(x) ({typeof(x + 0) y; y = x; y--; y;})

The trick is that the expression for typeof is x + 0, which is a r-value, and so the l-value-constness (which is what you want to avoid) is lost.

The same trick can be done with 1 * x, but curiously enough, +x and -x don't work.

Answered By - rodrigo

Answer Checked By - Terry (WPSolving Volunteer)

Thursday, April 7, 2022

[SOLVED] Is it possible to un-const typeof in gcc pure C?

Issue

Solution