Freigeben über


is_reference Class

Tests if type is a reference.

template<class Ty>
    struct is_reference;

Parameters

  • Ty
    The type to query.

Remarks

An instance of the type predicate holds true if the type Ty is a reference to an object or to a function, otherwise it holds false.

Example

 

// std__type_traits__is_reference.cpp 
// compile with: /EHsc 
#include <type_traits> 
#include <iostream> 
 
struct trivial 
    { 
    int val; 
    }; 
 
int main() 
    { 
    std::cout << "is_reference<trivial> == " << std::boolalpha 
        << std::is_reference<trivial>::value << std::endl; 
    std::cout << "is_reference<trivial&> == " << std::boolalpha 
        << std::is_reference<trivial&>::value << std::endl; 
    std::cout << "is_reference<int()> == " << std::boolalpha 
        << std::is_reference<int()>::value << std::endl; 
    std::cout << "is_reference<int(&)()> == " << std::boolalpha 
        << std::is_reference<int(&)()>::value << std::endl; 
 
    return (0); 
    } 
 
is_reference<trivial> == false
is_reference<trivial&> == true
is_reference<int()> == false
is_reference<int(&)()> == true

Requirements

Header: <type_traits>

Namespace: std

See Also

Reference

<type_traits>

is_pointer Class