On Wed, Nov 15, 2023 at 3:34 PM Frederick Virchanza Gotham via
Std-Proposals <std-proposals_at_[hidden]> wrote:

> On Wed, Nov 15, 2023 at 12:09 PM Pavel Vazharov <freakpv_at_[hidden]> wrote:
> > #include <cstdio>
> >
> > template <typename Tag = decltype([]{})>
> > void fun()
> > {
> > static int counter = 0;
> > std::printf("addr:%p value:%i\n", &counter, counter);
> > }
> >
> > int main()
> > {
> > fun();
> > fun();
> > fun();
> >
> > return 0;
> > }
> >
> > This prints something like
> >
> > addr:0x40401c value:0
> > addr:0x404020 value:0
> > addr:0x404024 value:0
>
> Does the Standard mandate that every C++23 compiler behaves this way?
> Or does it just so happen that this code behaves the way it behaves on
> all the major compilers? What I mean is: Could the compiler create the
> default template argument in one place and then share that type among
> the instantiations? Or, does the Standard mandate that the compiler
> produce a program that behaves as though:
>
> fun();
> fun();
> fun();
>
> had been written as:
>
> fun< decltype([]{}) >();
> fun< decltype([]{}) >();
> fun< decltype([]{}) >();
>
> With the way the C++23 Standard is currently written, could these
> three lines have the same type as the default template parameter,
> almost as though it had been written:
>
> auto some_static_duration_lambda_object = []{};
> fun< decltype(some_static_duration_lambda_object) >();
> fun< decltype(some_static_duration_lambda_object) >();
> fun< decltype(some_static_duration_lambda_object) >();
> --
>
As far as I know the standard mandates that every lambda has a unique type.
I don't think that the compiler is allowed to do the "collapsing" of the
"same" lambda objects to a single type but I can't prove it as I'm not so
versed with the C++ standard.