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Difference between revisions of "cpp/types/disjunction"

From cppreference.com
< cpp‎ | types
Line 36: Line 36:
 
  | 1=
 
  | 1=
 
template<class... B> struct disjunction
 
template<class... B> struct disjunction
     : integral_constant<bool, B::value {{!!}} ...>;
+
     : integral_constant<bool, (B::value {{!!}} ...)>;
 
}}
 
}}
  

Revision as of 02:52, 28 March 2017

 
 
Utilities library
General utilities
Relational operators (deprecated in C++20)
 
 
Defined in header <type_traits>
template<class... B>
struct disjunction;
(1) (since C++17)

Forms the logical disjunction of the type traits B..., effectively performing a logical or on the sequence of traits.

The specialization std::disjunction<B1, ..., BN> has a public and unambiguous base that is

  • if sizeof...(B) == 0, std::false_type; otherwise
  • the first type Bi in B1, ..., BN for which bool(Bi::value) == true, or BN if there is no such type.

The member names of the base class, other than disjunction and operator=, are not hidden and are unambiguously available in disjunction.

Disjunction is short-circuiting: if there is a template type argument Bi with bool(Bi::value) != false, then instantiating disjunction<B1, ..., BN>::value does not require the instantiation of Bj::value for j > i

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 disjunction_v = disjunction<B...>::value;
(since C++17)

Possible implementation

template<class... B> struct disjunction
    : integral_constant<bool, (B::value || ...)>;

Notes

A specialization of disjunction does not necessarily inherit from of either std::true_type or std::false_type: it simply inherits from the first B whose ::value, converted to bool, is true, or from the very last B when all of them convert to false. For example, std::disjunction<std::integral_constant<int, 2>, std::integral_constant<int, 4>>::value is 2.

Example

#include <iostream>
#include <type_traits>
 
using result0 =
    std::disjunction<std::bool_constant<false>, std::bool_constant<false>,
                     std::bool_constant<false>>;
using result1 =
    std::disjunction<std::bool_constant<true>, std::bool_constant<false>,
                     std::bool_constant<false>>;
 
int main()
{
    std::cout << std::boolalpha;
    std::cout << result0::value << '\n';
    std::cout << result1::value << '\n';
}

Output:

false
true

See also

(C++17)
logical NOT metafunction
(class template) [edit]
variadic logical AND metafunction
(class template) [edit]