On Fri, Nov 12, 2021, 8:53 AM Jason McKesson via Std-Discussion <
std-discussion_at_[hidden]> wrote:

> On Fri, Nov 12, 2021 at 3:26 AM Jens Maurer via Std-Discussion
> <std-discussion_at_[hidden]> wrote:
> >
> > On 12/11/2021 04.04, Hani Deek via Std-Discussion wrote:
> > > The following code is accepted by GCC, Clang and MSVC. I thought it
> was ill-formed because the variable V0 is not usable in constant
> expressions. Is this just a bug in all of those compilers? Or is there an
> explanation in the C++ rules?
> > >
> > > struct S{ constexpr operator int() const{ return 0; } };
> > >
> > > void foo(S V0) { constexpr int V1 = V0; }
> > >
> > > int main()
> > > {
> > > S V0{};
> > > foo(V0);
> > > }
> >
> > There is no lvalue-to-rvalue conversion on any part of V0.
> >
> > Do you feel any other provision in [expr.const] p5 should
> > apply, making this not a constant expression?
>
> I got a bit lost in the details, but my confusion comes from the use
> of the name `V0`. C++20 says:
>
> > An object or reference is usable in constant expressions if it is:
> > * a variable that is usable in constant expressions, or ...
>
> And:
>
> > A constant-initialized potentially-constant variable is usable in
> constant expressions ...
>
> This means that the only variables that are "usable in constant
> expressions" are those which are constant-initialized and potentially
> constant. Neither is true of `V0`.
>
> Are you suggesting that the use of the name `V0` does not trigger one
> of the conditions in [expr.const]/5 because it never triggers
> lvalue-to-rvalue conversion? If this is true, can you explain why it
> does not?
>

"Usable in constant expressions" is a specific term that is only referenced
in a few places. For instance, to invoke a virtual function on an object.
Or to access a reference (for now).

None of the operations in this example require V0 to be usable in constant
expressions. So the fact that it's not doesn't matter.

Barry

>