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Iterator: Use boost::result_of to determine nested result type of function in transform_iterator. Fixes #1427.

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[SVN r70715]
This commit is contained in:
Daniel James
2011-03-29 21:31:29 +00:00
parent 1ca1caddff
commit 5d72ae48da
2 changed files with 36 additions and 20 deletions
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25
include/boost/iterator/transform_iterator.hpp
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@ -20,6 +20,8 @@
#include <boost/type_traits/is_reference.hpp>
#include <boost/type_traits/remove_const.hpp>
#include <boost/type_traits/remove_reference.hpp>
#include <boost/utility/result_of.hpp>
#if BOOST_WORKAROUND(BOOST_MSVC, BOOST_TESTED_AT(1310))
# include <boost/type_traits/is_base_and_derived.hpp>
@ -35,33 +37,16 @@ namespace boost
namespace detail
{
template <class UnaryFunc>
struct function_object_result
{
typedef typename UnaryFunc::result_type type;
};
#ifndef BOOST_NO_TEMPLATE_PARTIAL_SPECIALIZATION
template <class Return, class Argument>
struct function_object_result<Return(*)(Argument)>
{
typedef Return type;
};
#endif
// Compute the iterator_adaptor instantiation to be used for transform_iterator
template <class UnaryFunc, class Iterator, class Reference, class Value>
struct transform_iterator_base
{
private:
// By default, dereferencing the iterator yields the same as
// the function. Do we need to adjust the way
// function_object_result is computed for the standard
// proposal (e.g. using Doug's result_of)?
// the function.
typedef typename ia_dflt_help<
Reference
, function_object_result<UnaryFunc>
, result_of<UnaryFunc(typename std::iterator_traits<Iterator>::reference)>
>::type reference;
// To get the default for Value: remove any reference on the
@ -113,7 +98,7 @@ namespace boost
#endif
}
template<
template <
class OtherUnaryFunction
, class OtherIterator
, class OtherReference