Need clarity on " v##a##r "execution?

Question

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define macro(r,a,v) v##a##r
#define DECVAR macro(r,a,v)

int main() {
int DECVAR=42;
var++;
printf("%d",var);
return 0;
}

ans=43

Answer 1

Have you tried checking the comment section of this feed post? In preprocessor commands, ## is concatenation. DECVAR is replaced with macro(r,a,v), which is then replaced with var. 42 is assigned to var, bar is incremented, and then printed.

https://www.sololearn.com/post/39626/?ref=app

Answer 2

Thanks sir.