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From: Maxim Yegorushkin (e-maxim_at_[hidden])
Date: 2004-06-10 03:20:09
Is there a way to make a boost::lambda functor adaptable, that is to make it define result_type? There is a ret<> function template that lets one specify the result type, but the returned functor still does not have result_type typedef.
I've encountered the problem trying to use boost::transform_iterator with a boost::lambda functor:
std::vector<int> intVect;
boost::make_transform_iterator(intVect.begin(), boost::lambda::_1 / 2);
transform_iterator<> does not compile comlaining the functor does not have result_type.
As a workaround I had to write something like:
template<class R, class T>
struct result_type_wrapper : T
{
typedef R result_type;
result_type_wrapper(T const& t) : T(t) {}
};
template<class R, class T>
inline
result_type_wrapper<R, T> result_type(T const& t) { return t; }
boost::make_transform_iterator(intVect.begin(), result_type<int>(boost::lambda::_1 / 2));
Am I missing something? Is there a way to avoid using result_type_wrapper<>?
-- Maxim Yegorushkin
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