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From: Ion Gaztañaga (igaztanaga_at_[hidden])
Date: 2021-08-07 21:29:07

On 06/08/2021 1:55, Gavin Lambert via Boost wrote:
> On 5/08/2021 8:09 pm, Andrey Semashev wrote:
>> On 8/5/21 4:18 AM, Gavin Lambert wrote:
>>>> No, not really. Not only because needing is_constant_evaluated does
>>>> not equate to needing type traits or metaprogramming, but because if
>>>> the macro check says "no" then you've included <type_traits> for
>>>> nothing.
>>> Yes, really.  That's the header in which std::is_constant_evaluated
>>> is defined (as well as the feature macro you're wanting to check), so
>>> if you're wanting to use it then you have to include the header anyway.
>>> There's not really any point in making the include itself conditional
>>> (even if you could), except perhaps on something more generic (such
>>> as "is C++11 or later", to detect cases when the header wouldn't exist).
>> The point is reducing compile times. When all you need from
>> <type_traits> is is_constant_evaluated, it is wasteful to include the
>> header, especially if that component doesn't exist.
> The probability that any real-world translation unit does not already
> include <type_traits> is approximately zero.  More than half the other
> STL headers implicitly include it anyway.  (Or at least some internal
> equivalent.)

Except for those object files not using the STL, which are a lot if you
want a decent compilation speed.


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