std::is_reference
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Defined in header
<type_traits> |
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template< class T >
struct is_reference; |
(since C++11) | |
If T is a reference type (lvalue reference or rvalue reference), provides the member constant value equal true. For any other type, value is false.
The behavior of a program that adds specializations for is_reference or is_reference_v (since C++17) is undefined.
Template parameters
| T | - | a type to check |
Helper variable template
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template< class T >
inline constexpr bool is_reference_v = is_reference<T>::value; |
(since C++17) | |
Inherited from std::integral_constant
Member constants
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value
[static]
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true if T is a reference type , false otherwise(public static member constant) |
Member functions
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operator bool
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converts the object to bool, returns value(public member function) |
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operator()
(C++14)
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returns value(public member function) |
Member types
| Type | Definition |
value_type |
bool |
type |
std::integral_constant<bool, value> |
Possible implementation
template <class T> struct is_reference : std::false_type {}; template <class T> struct is_reference<T&> : std::true_type {}; template <class T> struct is_reference<T&&> : std::true_type {}; |
Example
#include <iostream> #include <type_traits> class A {}; int main() { std::cout << std::boolalpha; std::cout << std::is_reference<A>::value << '\n'; std::cout << std::is_reference<A&>::value << '\n'; std::cout << std::is_reference<A&&>::value << '\n'; std::cout << std::is_reference<int>::value << '\n'; std::cout << std::is_reference<int&>::value << '\n'; std::cout << std::is_reference<int&&>::value << '\n'; }
Output:
false true true false true true
See also
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(C++11)
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checks if a type is a lvalue reference (class template) |
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(C++11)
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checks if a type is a rvalue reference (class template) |