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?