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 to 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.
The behavior of a program that adds specializations for any of the templates described on this page is undefined.
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) |
| value
[static] | true if T is assignable from U, 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> |
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.
#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
|
(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) |
|
(C++20) | specifies that a type is assignable from another type (concept) |
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