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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>
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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