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Subject: Re: [boost] [system] Would it be possible to trial a breaking change to Boost.System and see what happens?
From: Niall Douglas (s_sourceforge_at_[hidden])
Date: 2018-01-14 23:01:10

>> constexpr says that the compiler is allowed to assume that the
>> instance in the current compilation unit is the sole one in the
>> process. That lets it eliminate it. That's rather antithetical to the
>> hard requirements in the standard. This is why I've discounted your
>> LWG solution, without additional changes to permit multiple instances,
>> I don't think your example solution can work.
> `constexpr` can mean different things in different contexts. If you
> declare a constexpr variable of a literal type, yes, the compiler can
> eliminate it. That's not quite the case here though.
>    extern const __system_category_impl __syscat;
>    constexpr error_category const& system_category()
>    {
>        return __syscat;
>    }
> Here `__syscat` is neither constexpr nor is it of a literal type (it has
> a virtual destructor.)
> system_category() is constexpr because it returns a known symbolic
> address, &__syscat, which can be compared for equality with other known
> symbolic addresses, and the comparison is constexpr, that is, its result
> is a compile-time constant expression. If you try to do something at
> compile time with the actual value to which system_category() returned a
> reference though, you'll fail; the only compile-time constant part of it
> is the address.
> If you then declare
>    constexpr error_code ec( 0, system_category() );
> now that would be a constexpr variable of a literal type, and the
> compiler is allowed to assume that it's the same even if it appears
> twice, and is allowed to eliminate it. But that's another thing
> entirely, and does not affect whether __syscat is unique.
> To make this even more complicated, constexpr has a third meaning; when
> put on a constructor, it enables static initialization. This is what
> makes it possible for __syscat to be defined in a manner so that it's
> initialized before anything else.

I didn't believe you, so I wrote The key
is that the extern const variable is not a reference, but is a value.

Fair enough. You are correct. I have learned something new, and now I
need to go adjust some implementation code. Thank you.


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