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std::is_const
Defined in header <type_traits> |
||
|---|---|---|
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(since C++11) |
std::is_const is a UnaryTypeTrait.
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 std::is_const or std::is_const_v is undefined.
Template parameters
| T | - | a type to check |
Helper variable template
|
(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
|
Example
#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:
false
true
false because the pointer itself can be changed but not the int pointed at
true because the pointer itself can't be changed but the int pointed at can
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) |
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