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    | std::is_pointer_interconvertible_base_of(3) | C++ Standard Libary | std::is_pointer_interconvertible_base_of(3) | 
NAME¶
std::is_pointer_interconvertible_base_of - std::is_pointer_interconvertible_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 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.
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-qualification), 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_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
  
   Pointer-interconvertibility traits: __cpp_lib_is_pointer_interconvertible
    201907L (C++20) * std::is_pointer_interconvertible_base_of,
  
   * std::is_pointer_interconvertible_with_class
Example¶
// Run this code
  
   #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() {}
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)
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