std::function_ref::operator=
From cppreference.com
< cpp | utility | functional | function ref
| constexpr function_ref& operator=( const function_ref& ) noexcept = default; |
(1) | (since C++26) |
| template< class T > constexpr function_ref& operator=( T ) = delete; |
(2) | (since C++26) |
1) Copy assignment operator is explicitly-defaulted.
std::function_ref satisfies copyable and TriviallyCopyable. This defaulted assignment operator performs a shallow copy of the stored thunk-ptr and bound-entity.2) User-defined assignment operator is explicitly-deleted if T is not the same type as
std::function_ref, std::is_pointer_v<T> is false, and T is not a specialization of std::nontype_t. This overload participates in overload resolution only if the constraints are satisfied in the conditions above.Return value
*this
See also
constructs a new function_ref object (public member function) | |
| replaces or destroys the target (public member function of std::copyable_function) | |
| assigns a new target (public member function of std::function<R(Args...)>) | |
| replaces or destroys the target (public member function of std::move_only_function) |