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From: George Neuner <gneuner2@comcast.net>
Newsgroups: comp.compilers
Subject: Re: Preprocessing ##
Date: Mon, 15 May 2017 11:31:44 -0400
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On Sat, 13 May 2017 08:49:36 +0200, Hans-Peter Diettrich
<DrDiettrich1@netscape.net> wrote:

>How is the C preprocessor assumed to handle the ## concatenation operator?

It concatenate tokens *prior* to expansion.  If the token references a
parameter of the current macro, the (text of the) parameter is
substituted, otherwise the token is simply copied.

e.g, #define foo(x)  one ## two
     ... foo(help) =>  onetwo

     #define foo(x)  x ## _me
     ... foo(help) => help_me


>In the Windows headers I found macros like
>   #define foo(x) bar(##baz(x))
>where an expansion into ... (baz ... doesn't make sense, because "(baz"
>is not a valid preprocessor token. Can somebody explain the meaning and
>handling of such constructs?
>
>DoDi
>[I would guess that some macro between ( and ## expanded to
>nothing.  Or it could just be one of those text editing errors. -John]

I think John probably is correct that you've found typos.  If it were
the result of a macro substitution, you would see it only in the
output of the preprocessor - it would not be visible in an input
header file.

Moreover, although I'm not 100% certain, I believe the preprocessor
does not substitute inside (it's own) preprocessor definitions, but
rather only substitutes into C code.

George