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Groups > comp.compilers > #1884
| From | Hans-Peter Diettrich <DrDiettrich1@aol.com.dmarc.email> |
|---|---|
| Newsgroups | comp.compilers |
| Subject | Re: Preprocessing ## |
| Date | 2017-06-02 08:04 +0200 |
| Organization | Compilers Central |
| Message-ID | <17-06-003@comp.compilers> (permalink) |
| References | <17-05-003@comp.compilers> <17-05-006@comp.compilers> |
Am 15.05.2017 um 17:31 schrieb George Neuner: > 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. In a newer draft I found the constraint, that ## is an operator only if present in the replacement list (macro body), not from an argument. >> In the Windows headers I found macros like >> #define foo(x) bar(##baz(x)) Here the constraint is violated, ## is part of the argument and thus not recognized as a preprocessor operator. >> [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. In the meantime I guess that this construct is a creative use of the preprocessor, when *also* used for some other postprocessor. The C compiler may never see the ## if bar(x) has an empty definition, with the argument unused. This seems to be the case with above example, unless a project or tool adds global definitions which result in a different definition of bar(x). The standard also includes an example #define hash_hash # ## # which expands into a new ## token, which *is not* the ## operator. This may or may not mean that ## is recognized and handled as a preprocessor operator *only if* it occurs in the replacement list (body) of a macro definition. Such a token is not accepted subsequently by a C compiler, but may be recognized by some other tool. Likewise the standard requires that a # operator must be followed by a macro argument, else it is not recognized as a preprocessor operator. This signals to me that the preprocessor is allowed to output token sequences which are not valid C code, but may be valid when processed by some other tool. > 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. Right, immediate evaluation of # ## #, as a constant expression during macro definition, would result in a new ## token inside the replacement list, which would be recognized as another ## operator during a subsequent macro expansion. This were not the behaviour indicated by the description of above example. Thanks for your comments, without which I e.g. had guessed that constant expression evaluation were allowed during macro definition. DoDi
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Preprocessing ## Hans-Peter Diettrich <DrDiettrich1@netscape.net> - 2017-05-13 08:49 +0200
Re: Preprocessing ## Hans-Peter Diettrich <DrDiettrich1@netscape.net> - 2017-05-15 11:28 +0200
Re: Preprocessing ## Kaz Kylheku <686-678-9105@kylheku.com> - 2017-05-15 13:41 +0000
Re: Preprocessing ## George Neuner <gneuner2@comcast.net> - 2017-05-15 11:31 -0400
Re: Preprocessing ## Hans-Peter Diettrich <DrDiettrich1@aol.com.dmarc.email> - 2017-06-02 08:04 +0200
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