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
article.
The compiler gives an error message, but I can see valid C code by using -E
flag with the GCC compiler.
The macro X-list is defined as the following:
#define LIST \
X(red, "red") \
X(blue, "blue") \
X(yellow, "yellow")
And then:
#define X(a, b) foo.##a = -1;
LIST;
#undef X
But the gcc given the following errors messages:
lixo.c:42:1: error: pasting "." and "red" does not give a valid preprocessing token
lixo.c:42:1: error: pasting "." and "blue" does not give a valid preprocessing token
lixo.c:42:1: error: pasting "." and "yellow" does not give a valid preprocessing token
Like I said, I can seen valid C code by using -E
switch on gcc:
lixo.c:42:1: error: pasting "." and "red" does not give a valid preprocessing token
lixo.c:42:1: error: pasting "." and "blue" does not give a valid preprocessing token
lixo.c:42:1: error: pasting "." and "yellow" does not give a valid preprocessing token
foo.red = -1; foo.blue = -1; foo.yellow = -1;;
What's a valid preprocessing token? Can someone explain this?
(before you say "why not just an either initialize or memset()
?" it's not my real code.)
.
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
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