std::is_assignable,std::is_trivially_assignable,std::is_nothrow_assignable (3) - Linux Manuals

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

NAME

std::is_assignable,std::is_trivially_assignable,std::is_nothrow_assignable - std::is_assignable,std::is_trivially_assignable,std::is_nothrow_assignable

Synopsis

Defined in header <type_traits>
template< class T, class U > (1) (since C++11)
struct is_assignable;
template< class T, class U > (2) (since C++11)
struct is_trivially_assignable;
template< class T, class U > (3) (since C++11)
struct is_nothrow_assignable;

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.
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 > (since C++17)
inline constexpr bool is_assignable_v = is_assignable<T, U>::value;
template< class T, class U > (since C++17)
inline constexpr bool is_trivially_assignable_v = is_trivially_assignable<T, U>::value;
template< class T, class U > (since C++17)
inline constexpr bool is_nothrow_assignable_v = is_nothrow_assignable<T, U>::value;

Inherited from std::integral_constant

Member constants

value true if T is assignable from U , false otherwise
         (public static member constant)
[static]

Member functions

              converts the object to bool, returns value
operator bool (public member function)

operator() returns value
              (public member function)
(C++14)

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