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std::is_assignable,std::is_trivially_assignable,std::is_nothrow_assignable(3) C++ Standard Libary std::is_assignable,std::is_trivially_assignable,std::is_nothrow_assignable(3)

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


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 to 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.


The behavior of a program that adds specializations for any of the templates
described on this page 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 >
inline constexpr bool is_trivially_assignable_v = (since C++17)
is_trivially_assignable<T, U>::value;
template< class T, class U >
inline constexpr bool is_nothrow_assignable_v = (since C++17)
is_nothrow_assignable<T, U>::value;

Inherited from std::integral_constant


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


operator bool converts the object to bool, returns value
(public member function)
operator() returns value
(C++14) (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.

// 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';
}


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


is_copy_assignable
is_trivially_copy_assignable
is_nothrow_copy_assignable checks if a type has a copy assignment operator
(C++11) (class template)
(C++11)
(C++11)
is_move_assignable
is_trivially_move_assignable
is_nothrow_move_assignable checks if a type has a move assignment operator
(C++11) (class template)
(C++11)
(C++11)
assignable_from specifies that a type is assignable from another type
(C++20) (concept)

2022.07.31 http://cppreference.com

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