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").
Boost list run by bdawes at acm.org, gregod at cs.rpi.edu, cpdaniel at pacbell.net, john at johnmaddock.co.uk