Subject: Re: [boost] [bind] Function call operator is a template?!
From: Andy Venikov (avenikov_at_[hidden])
Date: 2009-12-20 12:12:26
Peter Dimov wrote:
> Mathias Gaunard wrote:
>> Andy Venikov a écrit :
>>
>>> I'm not going to debate whether it's right or wrong, but I'm still
>>> curious - why did it have to be a template? When bind() is called it
>>> already has all the information about the function's signature
>>
>> Not in the general case.
>> For function pointers or member function pointers, I suppose it is the
>> case though.
>
> It's not possible in general for function pointers either.
>
> void f( X x, Y y );
>
> bind( f, _1, _1 ); //?
Do you mean to say that there may be a type that converts to both X and Y,
but X and Y don't convert between each other?
Yup. There's a problem.
So it looks like templates are necessary.
But are unconstrained templates necessary? Couldn't we put a concept
check to enable templates only if underlying function/functor is
callable with the given arguments. Something to what Eric Niebler
proposed in another thread? ("How to detect if f() returns void").
Thanks,
Andy.