|Preprocessing ## DrDiettrich1@netscape.net (Hans-Peter Diettrich) (2017-05-13)|
|Re: Preprocessing ## DrDiettrich1@netscape.net (Hans-Peter Diettrich) (2017-05-15)|
|Re: Preprocessing ## firstname.lastname@example.org (Kaz Kylheku) (2017-05-15)|
|Re: Preprocessing ## email@example.com (George Neuner) (2017-05-15)|
|From:||George Neuner <firstname.lastname@example.org>|
|Date:||Mon, 15 May 2017 11:31:44 -0400|
|Organization:||A noiseless patient Spider|
|Injection-Info:||miucha.iecc.com; posting-host="news.iecc.com:2001:470:1f07:1126:0:676f:7373:6970"; logging-data="40555"; mail-complaints-to="email@example.com"|
|Posted-Date:||15 May 2017 15:23:54 EDT|
On Sat, 13 May 2017 08:49:36 +0200, Hans-Peter Diettrich
>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?
>[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
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.
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