std::experimental::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>

• if sizeof...(B) == 0, std::true_type
• the first type Bi in B1, ..., BN for which bool(Bi::value) == false , or BN

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

Helper variable template

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

#include <experimental/type_traits>
#include <iostream>
 
// func is enabled if all Ts... have the same type
template<typename T, typename... Ts>
constexpr std::enable_if_t<std::experimental::conjunction_v<std::is_same<T, Ts>...>>
func(T, Ts...)
{
    std::cout << "All types are the same.\n";
}
 
template<typename T, typename... Ts>
constexpr std::enable_if_t<!std::experimental::conjunction_v<std::is_same<T, Ts>...>>
func(T, Ts...)
{
    std::cout << "Types differ.\n";
}
 
int main()
{
    func(1, 2'7, 3'1);    
    func(1, 2.7, '3');    
}

All types are the same.
Types differ.

See also