Re: [Boost-users] Testing whether result_of<F(U)> is a valid expression

Subject: Re: [Boost-users] Testing whether result_of<F(U)> is a valid expression
From: TONGARI (tongari95_at_[hidden])
Date: 2013-02-08 08:14:55

2013/2/8 oswin krause <oswin.krause_at_[hidden]>

> Hi,
>
> Is it possible to evaluate, whether a certain argument type given to a
> functor will lead to a valid expression with the result_of template?
>
> i am trying to achieve the following, maybe there is a better way to do it:
>
> template<class T, class F>
> Container< typename boost::result_of<F(T)>::type > transform(
>     Container<T> c, F f,
>     std::disable_if<valid_**expression<F(Block<T>) > >::type* dummy = 0
> ){
>    //slow default implementation
>     Container<T> res(c.size());
>     std::transform(c.begin(),c.**end(),res.begin(),f):
>     return res;
> }
>
> //blocked implmentation of transform
> template<class T, class F>
> Container< typename boost::result_of<F(T)>::type > transform(
>     Container<T>, F functor,
>     std::enable_if<valid_**expression<F(Block<T>) > >::type* dummy = 0
> ){
> //functor supports fast implementation
>     Container<T> res(c.size());
> std::transform(c.blocks().**begin(),c.blocks().end(),res.**blocks().begin(),functor):
>
>     return res;
> }
>
> the question is, how valid_expression<F(Block<T>) > could be implemented
> and right now the only mechanism which get's close to what i want would be
> checking result_of<F(Block<T>)> is valid.
>

Hmm... I have implemented `has_call<F, R(Args..)>` some time ago, your code
may be rewritten as has_call<F, dont_care(Block<T>) >.

Interested?