RetroSearch Browse

Home - News ( United States | United Kingdom | Italy | Germany ) - Football scores

Showing content from https://en.cppreference.com/w/cpp/language/../../cpp/types/is_pointer_interconvertible_base_of.html below:

std::is_pointer_interconvertible_base_of - cppreference.com

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.

[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

U be a complete object type,
T be a complete object type with cv-qualification not less than U,
u be any valid lvalue of U,

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.

[edit] Example

#include <type_traits>
 
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() {}

[edit] See also

RetroSearch is an open source project built by @garambo | Open a GitHub Issue

Search and Browse the WWW like it's 1997 | Search results from DuckDuckGo

HTML: 3.2 | Encoding: UTF-8 | Version: 0.7.4