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.
Google search "boost can_be_called" or similar. Among other hits will probably be something Eric Niebler wrote for Boost.Proto (within whose documentation I think there's an explanation of how it works). TONGARI's implementation also pops up, I think.
- Jeff