On Fri, 23 Sept 2022 at 15:43, Jonathan Wakely wrote:
>
> On Fri, 23 Sept 2022 at 15:34, Marek Polacek wrote:
> >
> > On Thu, Sep 22, 2022 at 06:14:44PM -0400, Jason Merrill wrote:
> > > On 9/22/22 09:39, Marek Polacek wrote:
> > > > To improve compile times, the C++ library could use compiler built-ins
> > > > rather than implementing std::is_convertible (and _nothrow) as class
> > > > templates. This patch adds the built-ins. We already have
> > > > __is_constructible and __is_assignable, and the nothrow forms of those.
> > > >
> > > > Microsoft (and clang, for compatibility) also provide an alias called
> > > > __is_convertible_to. I did not add it, but it would be trivial to do
> > > > so.
> > > >
> > > > I noticed that our __is_assignable doesn't implement the "Access checks
> > > > are performed as if from a context unrelated to either type"
> > > > requirement,
> > > > therefore std::is_assignable / __is_assignable give two different
> > > > results
> > > > here:
> > > >
> > > > class S {
> > > > operator int();
> > > > friend void g(); // #1
> > > > };
> > > >
> > > > void
> > > > g ()
> > > > {
> > > > // #1 doesn't matter
> > > > static_assert(std::is_assignable<int&, S>::value, "");
> > > > static_assert(__is_assignable(int&, S), "");
> > > > }
> > > >
> > > > This is not a problem if __is_assignable is not meant to be used by
> > > > the users.
> > >
> > > That's fine, it's not.
> >
> > Okay then. libstdc++ needs to make sure then that it's handled right.
>
> That's fine, the type traits in libstdc++ are always "a context
> unrelated to either type", unless users do something idiotic like
> declare std::is_assignable as a friend.
>
> https://www.open-std.org/jtc1/sc22/wg21/docs/papers/2019/p1339r1.pdf
> wants to explicitly say that's idiotic.
And I just checked that a variable template like std::is_assignable_v
also counts as "a context unrelated to either type", even when
instantiated inside a member function of the type:
#include <type_traits>
template<typename T, typename U>
constexpr bool is_assignable_v = __is_assignable(T, U);
class S {
operator int();
friend void g(); // #1
};
void
g ()
{
// #1 doesn't matter
static_assert(std::is_assignable<int&, S>::value, "");
static_assert(std::is_assignable_v<int&, S>, "");
static_assert(__is_assignable(int&, S), "");
}
The first two assertions are consistent, and fail, which is what we
want. The direct use of the built-in succeeds, but we don't care.
