std::experimental::conjunction (3) - Linux Manuals

std::experimental::conjunction: std::experimental::conjunction

NAME

std::experimental::conjunction - std::experimental::conjunction

Synopsis

Defined in header <experimental/type_traits>
template<class... B> (library fundamentals TS v2)
struct conjunction;

Forms the logical_conjunction of the type traits B..., effectively performing a logical AND on the sequence of traits.
The specialization std::experimental::conjunction<B1, ..., BN> has a public and unambiguous base that is

* if sizeof...(B) == 0, std::true_type; otherwise
* the first type Bi in B1, ..., BN for which bool(Bi::value) == false, or BN if there is no such type.

The member names of the base class, other than conjunction and operator=, are not hidden and are unambiguously available in conjunction.
Conjunction is short-circuiting: if there is a template type argument Bi with bool(Bi::value) == false, then instantiating conjunction<B1, ..., BN>::value does not require the instantiation of Bj::value for j > i.

Template parameters

B... - every template argument Bi for which Bi::value is instantiated must be usable as a base class and define member value that is convertible to bool

Helper variable template

template<class... B> (library fundamentals TS v2)
constexpr bool conjunction_v = conjunction<B...>::value;

Possible implementation

  template<class...> struct conjunction : std::true_type { };
  template<class B1> struct conjunction<B1> : B1 { };
  template<class B1, class... Bn>
  struct conjunction<B1, Bn...>
      : std::conditional_t<bool(B1::value), conjunction<Bn...>, B1> {};

Notes

A specialization of conjunction does not necessarily inherit from either std::true_type or std::false_type: it simply inherits from the first B whose ::value, converted to bool, is false, or from the very last B when all of them convert to true. For example, conjunction<std::integral_constant<int, 2>, std::integral_constant<int, 4>>::value is 4.

Example

// Run this code

  // func is enabled if all Ts... have the same type
  template<typename T, typename... Ts>
  std::enable_if_t<std::experimental::conjunction_v<std::is_same<T, Ts>...> >
  func(T, Ts...) {
   // TODO something to show
  }