# std::is_assignable, std::is_trivially_assignable, std::is_nothrow_assignable

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< cpp‎ | types

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)

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

is_copy_assignable is_trivially_copy_assignable is_nothrow_copy_assignable (C++11)(C++11)(C++11) | checks if a type has a copy assignment operator (class template)
is_move_assignable is_trivially_move_assignable is_nothrow_move_assignable (C++11)(C++11)(C++11) | checks if a type has a move assignment operator (class template)

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