Re: Possible __VA_OPT__ bug

[Segher Boessenkool <segher@kernel.crashing.org>]

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