Subject: Re: [boost] [bind] Function call operator is a template?!
From: Mathias Gaunard (mathias.gaunard_at_[hidden])
Date: 2009-12-16 17:56:36
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.