On 10/31/2019 4:20 PM, Segher Boessenkool wrote:
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.
Clang in its C++20 mode treats __VA_OPT__ as I thought it should be,
where the determination of whether __VA_OPT__ ( pp-tokens ) expands to a
placeholder token or pp-tokens is determined by whether the arguments
for the variadic data parameter ... are empty or not after the arguments
have been fully macro replaced. In other words clang expands to 1 in
both my examples above.
Segher
