To support some compile time magic I would like to use pointers to members like:
struct BaseT
{
};
struct DerivedT: public BaseT
{
};
struct TestT
{
DerivedT testMem;
typedef BaseT (TestT::* TestTMemPtr);
constexpr TestT() = default;
static constexpr TestTMemPtr testMemOffset()
{
return &TestT::testMem;
}
};
int main()
{
constexpr TestT test;
}
I cannot return a pointer-to-derived-member as a pointer-to-base-member, I get this with clang:
cannot initialize return object of type 'TestT::TestTMemPtr' (aka 'BaseT (TestT::*)') with an rvalue of type 'DerivedT TestT::*'
I checked it with gcc:
error: invalid conversion from 'DerivedT TestT::*' to 'TestT::TestTMemPtr' {aka 'BaseT TestT::*'}
Is this the normal behavior? I thought I can always use a derived pointer as a base pointer.
UPDATE:
Ok, the original example wasn't the best, I think this one is more expressive, so DerivedT*
can be used as BaseT*
, but DerivedT TestT::*
cannot be used as BaseT TestT::*
:
struct BaseT
{
};
struct DerivedT: public BaseT
{
};
struct TestT
{
DerivedT m_test;
};
using BaseTMemPtr = BaseT TestT::*;
int main()
{
TestT test;
BaseT* simplePtr = &test.m_test; //It is DerivedT*, but can be used as BaseT*
BaseT (TestT::*memPtr) = &TestT::m_test; //Error, BaseT TestT::* cannot be used as DerivedT TestT::*
BaseTMemPtr memPtr2 = &TestT::m_test; //Error, just the same
}
CodePudding user response:
From the point of view of inheritance, BaseT TestT::*
and DerivedT TestT::*
are two unrelated types¹, so you can't initialize the former from the latter nor vice versa, just like you can't initialize a int*
with a double*
because int
and double
are not based and derived classes.
¹ By that I mean that two objects of these types don't point to two classes which are one the base of another. BaseT TestT::*
and DerivedT TestT::*
are both pointer types, but they don't point to two classes of which one is base of the other; they don't even point to classes in the first place (see demo code below), so there can be no inheritance relation between the pointed-to types, as inheritance is a thing between classes, not between types in general, such as member function types.
#include <type_traits>
struct BaseT { };
struct DerivedT: public BaseT { };
struct TestT { };
template<typename T, typename = void>
struct points_to_class : std::false_type {};
template<typename T>
struct points_to_class<T*> : std::is_class<T> {};
static_assert(points_to_class<BaseT*>::value); // passes
static_assert(points_to_class<BaseT TestT::*>::value); // fails
But so, is the conversion between pointers only possible when they both point to classes and those two classes are related by inheritance?
Well, if you give a look at the Pointer declaration page on cppreference.com, it does have a section on Pointers to member functions, and it is about conversion between pointers to member functions.
But it is about of a pointer to member function of a base class to pointer to the same member function of a derived class, whereas you seem to look for converting a pointer to member function (of TestT
) returning a base class (BaseT
) to a pointer to member function of the same class (TestT
) returning a derived class (DerivedT
). Again, the two types are unrelated.