std::conjunction - cppreference.com
From cppreference.com
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(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
BiinB1, ..., BNfor whichbool(Bi::value) == false, orBNif 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.
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
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Helper variable template
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(since C++17) | |
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, 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.
| Feature-test macro | Value | Std | Feature |
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__cpp_lib_logical_traits |
201510L |
(C++17) | Logical operator type traits |
Example
#include <iostream> #include <type_traits> // 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.