Std::is_copy_assignable,std::is_trivially_copy_assignable, (3) Linux Manual Page

std::is_copy_assignable,std::is_trivially_copy_assignable, – std::is_copy_assignable,std::is_trivially_copy_assignable,

Synopsis

Defined in header<type_traits>

    template <class T>
    (1)(since C++ 11)

        struct is_copy_assignable;

template <class T>
(2)(since C++ 11)

    struct is_trivially_copy_assignable;

template <class T>
(3)(since C++ 11)

    struct is_nothrow_copy_assignable;

1) If T is not a referenceable type (i.e., possibly cv-qualified void or a function
type with a cv-qualifier-seq or a ref-qualifier), provides a member constant value
equal to false. Otherwise, provides a member constant value equal to
std::is_assignable<T&, const T&>::value.
2) Same as 1), but uses std::is_trivially_assignable<T&, const T&>
3) Same as 1), but uses std::is_nothrow_assignable<T&, const T&>
T shall 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 >

inline constexpr bool is_copy_assignable_v =                           (since C++17)

is_copy_assignable<T>::value;

template< class T >

inline constexpr bool is_trivially_copy_assignable_v =                 (since C++17)

is_trivially_copy_assignable<T>::value;

template< class T >

inline constexpr bool is_nothrow_copy_assignable_v =                   (since C++17)

is_nothrow_copy_assignable<T>::value;

Inherited from std::integral_constant

Member constants

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

Member functions

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

Member types

Type Definition
value_type bool
type std::integral_constant<bool, value>

Possible implementation

template< class T> struct is_copy_assignable
: std::is_assignable< typename std::add_lvalue_reference<T>::type,
typename std::add_lvalue_reference<const T>::type> {};

template< class T> struct is_trivially_copy_assignable
: std::is_trivially_assignable< typename std::add_lvalue_reference<T>::type,
typename std::add_lvalue_reference<const T>::type> {};

template< class T> struct is_nothrow_copy_assignable
: std::is_nothrow_assignable< typename std::add_lvalue_reference<T>::type,
typename std::add_lvalue_reference<const T>::type> {};

Notes

The trait std::is_copy_assignable is less strict than CopyAssignable because it does
not check the type of the result of the assignment (which, for a CopyAssignable
type, must be an lvalue of type T) and does not check the semantic requirement that
the argument expression remains unchanged. It also does not check that T satisfies
MoveAssignable, which is required of all CopyAssignable types.

Example

// Run this code

#include <iostream>

#include <utility>

#include <type_traits>

struct Foo {
    int n;
};

int main()
{
    std::cout << std::boolalpha
              << "Foo is trivially copy-assignable? "
              << std::is_trivially_copy_assignable<Foo>::value << '\n'
              << "int[2] is copy-assignable? "
              << std::is_copy_assignable<int[2]>::value << '\n'
              << "int is nothrow copy-assignable? "
              << std::is_nothrow_copy_assignable<int>::value << '\n';
}

Output:

Foo is trivially copy-assignable? true
int[2] is copy-assignable? false
int is nothrow copy-assignable? true

std::is_copy_assignable,std::is_trivially_copy_assignable, (3) Linux Manual Page