#include <stdio.h>
#define f(a,b) a##b
#define g(a) #a
#define h(a) g(a)
int main()
{
printf("%s\n",h(f(1,2)));
printf("%s\n",g(f(1,2)));
return 0;
}
Just by looking at the program one "might" expect the output to be, the same for both the printf statements. But on running the program you get it as:
bash$ ./a.out
12
f(1,2)
bash$
Why is it so?
Because that is how the preprocessor works.
A single '#' will create a string from the given argument, regardless of what that argument contains, while the double '##' will create a new token by concatenating the arguments.
Try looking at the preprocessed output (for instance with gcc -E) if you want to understand better how the macros are evaluated.