On Thu, Oct 31, 2019 at 02:50:07PM -0400, Edward Diener wrote: > On 10/31/2019 1:14 PM, Segher Boessenkool wrote: > >On Thu, Oct 31, 2019 at 12:42:54PM -0400, Edward Diener wrote: > >>Given: > >> > >>#define NO_DATA > >>#define TRY_VA_OPT(...) __VA_OPT__ (0) 1 > >> > >>TRY_VA_OPT() -> expands to 1 as expected > >>TRY_VA_OPT(NO_DATA) -> expands to 0 1 which is not expected > >> > >>when compiled with gcc-9.2 with -std=c++2a. > > > >Why is that not expected? The variadic macro TRY_VA_OPT does get tokens > >in its variable argument (namely, NO_DATA), so __VA_OPT__ expands to its > >argument (which is 0). > > > >https://gcc.gnu.org/onlinedocs/cpp/Variadic-Macros.html > > The wording of 15.6.1 paragraph 3 is confusing to me. I had thought that > the arguments to the variadic ... parameter were completely macro > expanded before considering whether the __VA_OPT__ ( pp-tokens ) would > expand to 'pp_tokens' or a single placemarker token. Evidently you are > saying that the correct interpretation with the __VA_OPT__ construct is > that the variadic ... arguments are not macro expanded before > consideration of the __VA_OPT__ construct processing . This seems to me > very odd because the arguments to the variadic ... parameter are always > completely macro expanded before being replaced by any __VA_ARGS__ > parameter in the replacement list. I wonder why the C++ standard > committee decided to treat __VA_OPT__ differently from __VA_ARGS__ in > this regard ? I don't know. I don't know if the GCC implementation is correct, either. Segher
