std::conjunction - cppreference.com (original) (raw)

| | | | | ------------------------------------------- | | ------------- | | template< class... B > struct conjunction; | | (since C++17) |

Forms the logical conjunction of the type traits B..., effectively performing a logical AND on the sequence of traits.

The specialization std::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.

If the program adds specializations for std::conjunction or std::conjunction_v, the behavior is undefined.

Contents

• 1 Template parameters
• 2 Helper variable template
• 3 Possible implementation
• 4 Notes
• 5 Example
• 6 See also

[edit] Template parameters

[edit] Helper variable template

| template< class... B > constexpr bool conjunction_v = conjunction<B...>::value; | | (since C++17) | | ----------------------------------------------------------------------------------- | | ------------- |

[edit] Possible implementation

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

[edit] 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, explicitly converted to bool, is false, or from the very last B when all of them convert to true. For example, std::conjunction<std::integral_constant<int, 2>, std::integral_constant<int, 4>>::value is 4.

The short-circuit instantiation differentiates conjunction from fold expressions: a fold expression, like (... && Bs::value), instantiates every B in Bs, while std::conjunction_v<Bs...> stops instantiation once the value can be determined. This is particularly useful if the later type is expensive to instantiate or can cause a hard error when instantiated with the wrong type.

[edit] Example

#include #include   // func is enabled if all Ts... have the same type as T template<typename T, typename... Ts> std::enable_if_t<std::conjunction_v<std::is_same<T, Ts>...>> func(T, Ts...) { std::cout << "All types in pack are the same.\n"; }   // otherwise template<typename T, typename... Ts> std::enable_if_t<!std::conjunction_v<std::is_same<T, Ts>...>> func(T, Ts...) { std::cout << "Not all types in pack are the same.\n"; }   template<typename T, typename... Ts> constexpr bool all_types_are_same = std::conjunction_v<std::is_same<T, Ts>...>;   static_assert(all_types_are_same<int, int, int>); static_assert(not all_types_are_same<int, int&, int>);   int main() { func(1, 2, 3); func(1, 2, "hello!"); }

Output:

All types in pack are the same. Not all types in pack are the same.

[edit] See also