# std::experimental::conjunction

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< cpp‎ | experimental

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

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

### 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

    #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');
    }

Output:

    All types are the same.
    Types differ.

### See also

conjunction (C++17) | variadic logical AND metafunction (class template)

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