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.[edit]Return value
*this
[edit]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 ) |