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)  | |
|   assigns a new target  (public member function of std::function<R(Args...)>)  |