Re: [boost] New, powerful way to use enable_if in C++0x

Subject: Re: [boost] New, powerful way to use enable_if in C++0x
From: Gevorg Voskanyan (v_gevorg_at_[hidden])
Date: 2011-04-11 17:16:26

Matt Calabrese wrote:
> Gevorg Voskanyan wrote:
> > Is  there a particular reason
> > you
> > did it like:
> > typename  boost::enable_if< boost::is_arithmetic< T > >::type*& =
> >  boost::enabler
> > instead of:
> > typename Enable = typename  boost::enable_if< boost::is_arithmetic< T >
> > >::type
> >  ?
> > The latter seems clearer to me and removes the need to introduce  a
> > boost::enabler identifier.
> 
> 
> It's necessary because otherwise  with two overloads you'd end up with two
> declarations that differ only in  default arguments, which would be a compile
> error.

Yes, of course that's very true, I wonder how could I miss that!
This could be included in the rationale section of the doc to prevent other 
users from being tempted to "simplify" it as I was. :)

[snip]

Thanks,
Gevorg