error: pasting “.” and “red” does not give a valid preprocessing token

Question

I\'m implementing The X macro, but I have a problem with a simple macro expansion. This macro (see below) is used into several macros usage examples, by including in this articl

Answer 1

. separates tokens and so you can't use ## as .red is not a valid token. You would only use ## if you were concatenating two tokens into a single one.

This works:

#define X(a, b) foo.a = -1;

What's a valid proprocessing token? Can someone explain this?

It is what gets parsed/lexed. foo.bar would be parsed as 3 tokens (two identifiers and an operator): foo . bar If you use ## you would get only 2 tokens (one identifier and one invalid token): foo .bar