std::is_assignable, std::is_trivially_assignable, std::is_nothrow_assignable
Defined in header <type_traits> | ||
---|---|---|
template< class T, class U > struct is_assignable; | (1) | (since C++11) |
template< class T, class U > struct is_trivially_assignable; | (2) | (since C++11) |
template< class T, class U > struct is_nothrow_assignable; | (3) | (since C++11) |
std::declval<T>() = std::declval<U>()
is well-formed in unevaluated context, provides the member constant value
equal true
. Otherwise, value
is false
. Access checks are performed as if from a context unrelated to either type. std::declval
is considered trivial and not considered an odr-use of std::declval
.T
and U
shall each be a complete type, (possibly cv-qualified) void
, or an array of unknown bound. Otherwise, the behavior is undefined.
If an instantiation of a template above depends, directly or indirectly, on an incomplete type, and that instantiation could yield a different result if that type were hypothetically completed, the behavior is undefined.
Helper variable templates
template< class T, class U > inline constexpr bool is_assignable_v = is_assignable<T, U>::value; | (since C++17) | |
template< class T, class U > inline constexpr bool is_trivially_assignable_v = is_trivially_assignable<T, U>::value; | (since C++17) | |
template< class T, class U > inline constexpr bool is_nothrow_assignable_v = is_nothrow_assignable<T, U>::value; | (since C++17) |
Inherited from std::integral_constant
Member constants
value
[static] | true if T is assignable from U , 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
This trait does not check anything outside the immediate context of the assignment expression: if the use of T
or U
would trigger template specializations, generation of implicitly-defined special member functions etc, and those have errors, the actual assignment may not compile even if std::is_assignable<T,U>::value
compiles and evaluates to true
.
Example
#include <iostream> #include <string> #include <type_traits> struct Ex1 { int n; }; int main() { std::cout << std::boolalpha << "int is assignable from int? " << std::is_assignable<int, int>::value << '\n' // 1 = 1; wouldn't compile << "int& is assignable from int? " << std::is_assignable<int&, int>::value << '\n' // int a; a = 1; works << "int is assignable from double? " << std::is_assignable<int, double>::value << '\n' << "int& is nothrow assignable from double? " << std::is_nothrow_assignable<int&, double>::value << '\n' << "string is assignable from double? " << std::is_assignable<std::string, double>::value << '\n' << "Ex1& is trivially assignable from const Ex1&? " << std::is_trivially_assignable<Ex1&, const Ex1&>::value << '\n'; }
Output:
int is assignable from int? false int& is assignable from int? true int is assignable from double? false int& is nothrow assignable from double? true string is assignable from double? true Ex1& is trivially assignable from const Ex1&? true
See also
(C++11)(C++11)(C++11) | checks if a type has a copy assignment operator (class template) |
(C++11)(C++11)(C++11) | checks if a type has a move assignment operator (class template) |
© cppreference.com
Licensed under the Creative Commons Attribution-ShareAlike Unported License v3.0.
http://en.cppreference.com/w/cpp/types/is_assignable