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Explicitly deduced return type (like "auto foo() -> auto"), is that valid?
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Alf P. Steinbach
Aug 20, 2019, 10:59:56 PM8/20/19
to
Specifically, is
auto foo() -> auto { return 42; }
auto main() -> int
{
return foo();
}
... valid?
I'd guess that it's not valid as C++11, but possibly valid as C++14 or
later.
It compiles with Visual C++ 2019 and g++ 9.2.
Cheers!,
- Alf
auto foo() -> auto { return 42; }
auto main() -> int
{
return foo();
}
... valid?
I'd guess that it's not valid as C++11, but possibly valid as C++14 or
later.
It compiles with Visual C++ 2019 and g++ 9.2.
Cheers!,
- Alf
Tim Rentsch
Aug 22, 2019, 8:05:19 AM8/22/19
to
"Alf P. Steinbach" <alf.p.stein...@gmail.com> writes:
> Specifically, is
>
> auto foo() -> auto { return 42; }
>
> auto main() -> int
> {
> return foo();
> }
>
> ... valid?
>
> I'd guess that it's not valid as C++11, but possibly valid as
> C++14 or later.
fwiw, g++ and clang accept it without complaint under -std=c++14
> Specifically, is
>
> auto foo() -> auto { return 42; }
>
> auto main() -> int
> {
> return foo();
> }
>
> ... valid?
>
> I'd guess that it's not valid as C++11, but possibly valid as
> C++14 or later.
and -std=c++17, but give a diagnostic under -std=c++11. That
result holds both with and without a -pedantic option.
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