templated lambda issue

[Bonita Montero]

int main()
{
auto f = []<typename T>() -> T
{
return 123;
};
f.operator ()<int>(); // only way ?
}

Is the above code the only way to call the lambda ?

[David Brown]

You made a templated lambda - you have to give the type /somehow/. What
are you actually trying to do?

[Bonita Montero]

>> int main()
>> {
>>     auto f = []<typename T>() -> T
>>     {
>>         return 123;
>>     };
>>     f.operator ()<int>(); // only way ?
>> }
>>
>> Is the above code the only way to call the lambda ?

> You made a templated lambda - you have to give the type /somehow/.

No, I just want to return a value of that type.

[Alf P. Steinbach]

I've never used a templated lambda, but if I needed something like that
I think I'd use a type-indicating argument, like (off the cuff)

template< class T > struct Use_type_{ using Type = T; };

auto main() -> int
{
const auto f = []( auto t )
{ return static_cast<decltype(t)::Type>( 123 ); };

return f( Use_type_<int>() );
}

- Alf

[MrSpook...@qvqo7xsmj4s.co.uk]

On Mon, 14 Jun 2021 15:12:37 +0200
"Alf P. Steinbach" <alf.p.s...@gmail.com> wrote:
>On 14 Jun 2021 13:10, Bonita Montero wrote:
>> int main()
>> {
>>     auto f = []<typename T>() -> T
>>     {
>>         return 123;
>>     };
>>     f.operator ()<int>(); // only way ?
>> }
>>
>> Is the above code the only way to call the lambda ?
>
>I've never used a templated lambda, but if I needed something like that
>I think I'd use a type-indicating argument, like (off the cuff)
>
> template< class T > struct Use_type_{ using Type = T; };
>
> auto main() -> int

Or back in the real world: int main()