C preprocessor: stringize macro and identity macro

Question

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         


        

Answer 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)".