Re: [boost] [local] Help for the Alternatives section

Subject: Re: [boost] [local] Help for the Alternatives section
From: Mathias Gaunard (mathias.gaunard_at_[hidden])
Date: 2011-05-08 11:22:29

On 08/05/2011 16:04, Lorenzo Caminiti wrote:

>> auto /* or the real type */ l = overload(l1, l2);
>
> Does Boost already have a functor overloader like overload() above?

I do not think so.

> template<typename F0, typename F1, typename F2 = void, typename F3 = void>
> struct overload {
> };
>
> template<typename F0R, typename F0A0, typename F1R, typename F1A0>
> struct overload<F0R (F0A0), F1R (F1A0)>  {
>      overload(boost::function<F0R (F0A0)>  f0, boost::function<F1R (F1A0)>  f1):
>              f0_(f0), f1_(f1) {}
>      F0R operator()(F0A0 a0) const { return f0_(a0); }
>      F1R operator()(F1A0 a0) const { return f1_(a0); }
> private:
>      boost::function<F0R (F0A0)>  f0_;
>      boost::function<F1R (F1A0)>  f1_;
> };
>
> // More specializations to overload also with F2, F3, etc.

The type erasure appear unnecessary, and this implementation has big 
forwarding problems.

What about

template<typename F0, typename F1>
struct overload_t : F0, F1
{
     overload_t(F0 const& f0, F1 const& f1) : F0(f0), F1(f1)
     {
     }

     using F0::operator();
     using F1::operator();
};

template<typename F0, typename F1>
overload_t<F0, F1> overload(F0 const& f0, F1 const& f1)
{
     return overload_t<F0, F1>(f0, f1);
}

Function pointers will have to be dealt with specially, though.