std::void_t
From cppreference.com
Defined in header
<type_traits>
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template< class... >
using void_t = void; |
(since C++17) | |
Utility metafunction that maps a sequence of any types to the type void
[edit] Notes
This metafunction is used in template metaprogramming to detect ill-formed types in SFINAE context:
// primary template handles types that have no nested ::type member: template< class, class = std::void_t<> > struct has_type_member : std::false_type { }; // specialization recognizes types that do have a nested ::type member: template< class T > struct has_type_member<T, std::void_t<typename T::type>> : std::true_type { };
It can also be used to detect validity of an expression:
// primary template handles types that do not support pre-increment: template< class, class = std::void_t<> > struct has_pre_increment_member : std::false_type { }; // specialization recognizes types that do support pre-increment: template< class T > struct has_pre_increment_member<T, std::void_t<decltype( ++std::declval<T&>() )> > : std::true_type { };
Until C++17, unused parameters in alias templates were not guaranteed to ensure SFINAE and could be ignored, so pre-C++17 compilers require a more complex definition of void_t
, such as
template<typename... Ts> struct make_void { typedef void type;}; template<typename... Ts> using void_t = typename make_void<Ts...>::type;
[edit] Examples
This section is incomplete Reason: no example |
[edit] See also
(C++11)
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hides a function overload or template specialization based on compile-time boolean (class template) |