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templated lambda issue

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  templated lambda issue Bonita Montero <Bonita.Montero@gmail.com> - 2021-06-14 13:10 +0200
    Re: templated lambda issue David Brown <david.brown@hesbynett.no> - 2021-06-14 14:23 +0200
      Re: templated lambda issue Bonita Montero <Bonita.Montero@gmail.com> - 2021-06-14 14:58 +0200
    Re: templated lambda issue "Alf P. Steinbach" <alf.p.steinbach@gmail.com> - 2021-06-14 15:12 +0200
      Re: templated lambda issue MrSpook_5fdc40@qvqo7xsmj4s.co.uk - 2021-06-14 13:16 +0000

#80392 — templated lambda issue

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 ?

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#80393

On 14/06/2021 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 ?

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

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#80395

>> 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.

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#80397

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

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

- Alf

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#80398

On Mon, 14 Jun 2021 15:12:37 +0200
"Alf P. Steinbach" <alf.p.steinbach@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()

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