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 to true
. For any other type, value
is false
.
The behavior of a program that adds specializations for is_const
or is_const_v
(since C++17) is undefined.
T | - | a type to check |
template< class T > inline constexpr bool is_const_v = is_const<T>::value; | (since C++17) |
value
[static] | true if T is a const-qualified type, false otherwise (public static member constant) |
operator bool | converts the object to bool, returns value (public member function) |
operator()
(C++14) | returns value (public member function) |
Type | Definition |
---|---|
value_type | bool |
type | std::integral_constant<bool, value> |
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>
.
#include <iostream> #include <type_traits> int main() { std::cout << std::boolalpha << std::is_const_v<int> << '\n' // false << std::is_const_v<const int> << '\n' // true << std::is_const_v<const int*> // false << " because the pointer itself can be changed but not the int pointed at\n" << std::is_const_v<int* const> // true << " because the pointer itself can't be changed but the int pointed at can\n" << std::is_const_v<const int&> << '\n' // false << std::is_const_v<std::remove_reference_t<const int&>> << '\n' // true ; }
Output:
(C++11) | checks if a type is volatile-qualified (class template) |
(C++17) | obtains a reference to const to its argument (function template) |
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