std::function_ref::operator=
From cppreference.com
constexpr function_ref& operator=( const function_ref& ) noexcept = default;
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(1) | (since C++26) |
template< class T >
constexpr function_ref& operator=( T ) = delete;
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(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)
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| assigns a new target (public member function of std::function<R(Args...)>)
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| replaces or destroys the target (public member function of std::move_only_function)
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