std::is_assignable, std::is_trivially_assignable, std::is_nothrow_assignable
From cppreference.com
Defined in header <type_traits>
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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) |
1) If the expression 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. 2) same as (1), but the evaluation of the assignment expression will not call any operation that is not trivial. For the purposes of this check, a call to std::declval is considered trivial and not considered an odr-use of std::declval.
3) same as (1), but the evaluation of the assignment expression will not call any operation that is not noexcept.
T
and U
shall each be a complete type, (possibly cv-qualified) void, or an array of unknown bound. Otherwise, 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
Run this code
#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) |