std::is_pointer_interconvertible_base_of - cppreference.com (original) (raw)
| | | | | -------------------------------------------------------------------------------------- | | ------------- | | template< class Base, class Derived > struct is_pointer_interconvertible_base_of; | | (since C++20) |
If Derived is unambiguously derived from Base and every Derived object is pointer-interconvertible with its Base subobject, or if both are the same non-union class (in both cases ignoring cv-qualification), provides the member constant value equal to true. Otherwise value is false.
If both Base and Derived are non-union class types, and they are not the same type (ignoring cv-qualification), Derived shall be a complete type; otherwise the behavior is undefined.
If the program adds specializations for std::is_pointer_interconvertible_base_of or std::is_pointer_interconvertible_base_of_v, the behavior is undefined.
Contents
[edit] Helper variable template
| template< class Base, class Derived > inline constexpr bool is_pointer_interconvertible_base_of_v = is_pointer_interconvertible_base_of<Base, Derived>::value; | | (since C++20) | | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | | ------------- |
Inherited from std::integral_constant
Member constants
| | true if Derived is unambiguously derived from Base and every Derived object is pointer-interconvertible with its Base subobject, or if both are the same non-union class (in both cases ignoring cv-qualification), false otherwise (public static member constant) | | ------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------ |
Member functions
| | converts the object to bool, returns value (public member function) | | ---------------------------------------------------------------------- | | | returns value (public member function) |
Member types
[edit] Notes
std::is_pointer_interconvertible_base_of_v<T, U> may be true even if T is a private or protected base class of U.
Let
Ube a complete object type,Tbe a complete object type with cv-qualification not less thanU,ube any valid lvalue ofU,
reinterpret_cast<T&>(u) always has well-defined result if std::is_pointer_interconvertible_base_of_v<T, U> is true.
If T and U are not the same type (ignoring cv-qualification) and T is a pointer-interconvertible base class of U, then both std::is_standard_layout_v<T> and std::is_standard_layout_v<U> are true.
If T is standard layout class type, then all base classes of T (if any) are pointer-interconvertible base class of T.
| Feature-test macro | Value | Std | Feature |
|---|---|---|---|
| __cpp_lib_is_pointer_interconvertible | 201907L | (C++20) | Pointer-interconvertibility traits: std::is_pointer_interconvertible_base_of, std::is_pointer_interconvertible_with_class |
[edit] Example
#include struct Foo {}; struct Bar {}; class Baz : Foo, public Bar { int x; }; class NonStdLayout : public Baz { int y; }; static_assert(std::is_pointer_interconvertible_base_of_v<Bar, Baz>); static_assert(std::is_pointer_interconvertible_base_of_v<Foo, Baz>); static_assert(not std::is_pointer_interconvertible_base_of_v<Baz, NonStdLayout>); static_assert(std::is_pointer_interconvertible_base_of_v<NonStdLayout, NonStdLayout>); int main() {}