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map: Switch to SFINAE in template parameters

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Fixes #163.

Since C++11 we can do SFINAE in template parameters. VS2017 15.5 doesn't like
the SFINAE on argument here, but do not ask me why, because I have no idea.
This commit is contained in:
Nikita Kniazev
2017-12-13 23:36:08 +03:00
parent 821aaee960
commit 099333b61d
1 changed files with 6 additions and 9 deletions
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15
include/boost/fusion/container/map/map.hpp
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@ -67,24 +67,21 @@ namespace boost { namespace fusion
: base_type(std::forward<map>(seq))
{}
template <typename Sequence>
template <typename Sequence, typename = typename enable_if<traits::is_sequence<Sequence>>::type>
BOOST_FUSION_GPU_ENABLED
map(Sequence const& seq
, typename enable_if<traits::is_sequence<Sequence>, detail::enabler_>::type = detail::enabler)
map(Sequence const& seq)
: base_type(begin(seq), detail::map_impl_from_iterator())
{}
template <typename Sequence>
template <typename Sequence, typename = typename enable_if<traits::is_sequence<Sequence>>::type>
BOOST_FUSION_GPU_ENABLED
map(Sequence& seq
, typename enable_if<traits::is_sequence<Sequence>, detail::enabler_>::type = detail::enabler)
map(Sequence& seq)
: base_type(begin(seq), detail::map_impl_from_iterator())
{}
template <typename Sequence>
template <typename Sequence, typename = typename enable_if<traits::is_sequence<Sequence>>::type>
BOOST_FUSION_GPU_ENABLED
map(Sequence&& seq
, typename enable_if<traits::is_sequence<Sequence>, detail::enabler_>::type = detail::enabler)
map(Sequence&& seq)
: base_type(begin(seq), detail::map_impl_from_iterator())
{}