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cc-preprocessorstringification

How does this C code work?

What is a##b & #a?

  #define f(a,b) a##b
  #define g(a)   #a
  #define h(a) g(a)

  main()
  {
          printf("%s\n",h(f(1,2)));  //how should I interpret this?? [line 1]
          printf("%s\n",g(f(1,2)));  //and this? [line 2]
  }

How does this program work?

The output is

12
f(1, 2)

now I understand how a##b & #a work. But why is the result different in the two cases (line 1 and line 2)?

Solution

  • The ## concatenates two tokens together. It can only be used in the preprocessor.

    f(1,2) becomes 1 ## 2 becomes 12.

    The # operator by itself stringifies tokens: #a becomes "a". Therefore, g(f(1,2)) becomes "f(1,2)" when the preprocessor is done with it.

    h(f(1,2)) is effectively #(1 ## 2) which becomes #12 which becomes "12" as the preprocessor runs over it.