Difference between revisions of "cpp/types/conjunction"
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Conjunction is short-circuiting: if there is a template type argument {{tt|Bi}} with {{c|1= bool(Bi::value) == false}}, then instantiating {{c|conjunction<B1, ..., BN>::value}} does not require the instantiation of {{c|Bj::value}} for {{tt|j > i}}. | Conjunction is short-circuiting: if there is a template type argument {{tt|Bi}} with {{c|1= bool(Bi::value) == false}}, then instantiating {{c|conjunction<B1, ..., BN>::value}} does not require the instantiation of {{c|Bj::value}} for {{tt|j > i}}. | ||
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===Template parameters=== | ===Template parameters=== | ||
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Revision as of 09:04, 25 January 2020
Defined in header <type_traits>
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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
inB1, ..., BN
for which bool(Bi::value) == false, orBN
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 |
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> inline constexpr bool conjunction_v = conjunction<B...>::value; |
(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.
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 T\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 T\n"; } int main() { func(1, 2, 3); func(1, 2, "hello!"); }
Output:
all types in pack are T not all types in pack are T
See also
(C++17) |
logical NOT metafunction (class template) |
(C++17) |
variadic logical OR metafunction (class template) |