C preprocessor: stringize macro and identity macro

问题:

I want to know the reason behind the output of this code. I couldn't come up with an answer.

#define f(a,b) a##b
#define g(a) #a
#define h(a) g(a)
void main()
{
   printf("%s %s",h(f(1,2)),g(f(1,2)));
}

PS: output is 12 f(1,2). I thought it was 12 12 or f(1,2) f(1,2).

回答1:

h(f(1,2))

f(1,2) is substituted for a. a is not the subject of a # or ## operator so it's expanded to 12. Now you have g(12) which expands to "12".

g(f(1,2))

f(1,2) is substituted for a. The # operator applied to a prevents macro expansion, so the result is literally "f(1,2)".

回答2:

Just do the replacements.

h(f(1, 2)) -> g(12) -> "12"

g(f(1,2)) -> "f(1, 2)"

You should also see here.