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std::is_poi...ble_base_of(3) C++ Standard Libary std::is_poi...ble_base_of(3) NAME std::is_pointer_interconvertible_base_of - std::is_pointer_intercon- vertible_base_of Synopsis Defined in header <type_traits> template< class Base, class Derived > (since C++20) struct is_pointer_interconvertible_base_of; If Derived is unambiguously derived from Base and every Derived ob- ject 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; oth- erwise the behavior is undefined. The behavior of a program that adds specializations for is_pointer_interconvertible_base_of or is_pointer_interconvert- ible_base_of_v is undefined. Helper variable template template< class Base, class Derived > inline constexpr bool is_pointer_interconvertible_base_of_v = (since C++20) is_pointer_interconvertible_base_of<Base, Derived>::value; Inherited from std::integral_constant Member constants true if Derived is unambiguously derived from Base and every Derived object value is pointer-interconvertible with its Base subobject, or if both are the [static] same non-union class (in both cases ignoring cv-qualifica- tion), false otherwise (public static member constant) Member functions operator bool converts the object to bool, returns value (public member function) operator() returns value (C++14) (public member function) Member types Type Definition value_type bool type std::integral_constant<bool, value> 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_stan- dard_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: __cpp_lib_is_pointer_interconvertible Example // Run this code #include <iostream> #include <type_traits> struct Foo {}; struct Bar {}; class Baz : Foo, public Bar { int x; }; class NonStdLayout : public Baz { int y; }; int main() { std::cout << std::boolalpha << std::is_pointer_interconvertible_base_of_v<Bar, Baz> << '\n' << std::is_pointer_interconvertible_base_of_v<Foo, Baz> << '\n' << std::is_pointer_interconvertible_base_of_v<Baz, NonStdLay- out> << '\n' << std::is_pointer_interconvertible_base_of_v<NonStdLayout, NonStdLayout> << '\n'; } Output: true true false true See also is_base_of checks if a type is derived from the other type (C++11) (class template) is_empty checks if a type is a class (but not union) type and has no (C++11) non-static data members (class template) is_standard_layout checks if a type is a standard-layout type (C++11) (class template) http://cppreference.com 2022.07.31 std::is_poi...ble_base_of(3)
NAME | Synopsis | Member constants | Member functions | Member types | Notes | Example | Output: | See also
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