# and ## in macros

Question

  #include 
  #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(         


        

Answer 1

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.