std::is_const
From cppreference.com
| Defined in header <type_traits>
|
||
| template< class T > struct is_const; |
(since C++11) | |
If T is a const-qualified type (that is, const, or const volatile), provides the member constant value equal true. For any other type, value is false.
Template parameters
| T | - | a type to check |
Helper variable template
| template< class T > inline constexpr bool is_const_v = is_const<T>::value; |
(since C++17) | |
Inherited from std::integral_constant
Member constants
| value [static] |
true if T is a const-qualified type , false otherwise (public static member constant) |
Member functions
| operator bool |
converts the object to bool, returns value (public member function) |
| operator() (C++14) |
returns value (public member function) |
Member types
| Type | Definition |
value_type
|
bool
|
type
|
std::integral_constant<bool, value> |
Notes
If T is a reference type then is_const<T>::value is always false. The proper way to check a potentially-reference type for const-ness is to remove the reference: is_const<typename remove_reference<T>::type>.
Possible implementation
template<class T> struct is_const : std::false_type {}; template<class T> struct is_const<const T> : std::true_type {}; |
Example
Run this code
#include <iostream> #include <type_traits> int main() { std::cout << std::boolalpha; std::cout << std::is_const<int>::value << '\n'; // false std::cout << std::is_const<const int>::value << '\n'; // true std::cout << std::is_const<const int*>::value << '\n'; // false std::cout << std::is_const<int* const>::value << '\n'; // true std::cout << std::is_const<const int&>::value << '\n'; // false std::cout << std::is_const<typename std::remove_reference<const int&>::type>::value << '\n'; // true }
Output:
false true false true false true
See also
| (C++11) |
checks if a type is volatile-qualified (class template) |
| (C++17) |
obtains a reference to const to its argument (function template) |