Accessing base class member variable address from derived class when both have the same variable name

Question

I am trying to get the address of a base class protected member variable.

Please help me understand what's wrong with this code:

class A { protected: int m_; }; class B : public A { public: void Foo() { auto p1 = m_; // works but not what I want (B's member) auto p2 = &m_; // works but not the the address which I'm looking for (B's member) auto p3 = A::m_; // works but still not the address of the variable (A's member but not the address) auto p4 = &A::m_; // doesn't work -> int A::m_’ is protected within this context auto p5 = static_cast<A*>(this)->m_; // doesn't work -> int A::m_’ is protected within this context auto p6 = &static_cast<A*>(this)->m_; // doesn't work -> int A::m_’ is protected within this context } private: int m_; };

Thanks

Gil

JaMiT · Accepted Answer · 2020-03-28 01:54:11Z

A problem you ran into is that auto p4 = &A::m_; declares a pointer to data member. That is, the auto resolved itself to int A::*, not the int * you had intended. You can get the intended pointer by adding parentheses:

auto p4 = &(A::m_);

or if you want to make sure you get the type you intend:

int * p4 = &(A::m_);

Miles Budnek · Accepted Answer · 2020-03-28 02:00:19Z

&static_cast<A*>(this)->m_ doesn't work because access to protected base class members is only allowed via a (reference or pointer to an) object of the same type as the class that the accessing function is a member (or friend) of.

That is, the following is not valid, even though A is a base of B

class B : public A { public: void Foo(const A& a) { auto i = a.m_; // Not valid } };

See [class.protected].

static_cast<A*>(this) creates a pointer to A, which B cannot use to access A's protected members.

The issue with &A::m_ is that it's not taking a pointer to the A::m_ member of the current object; it's actually forming a pointer-to-member. i.e.

struct A { int m_; } int main() { auto foo = &A::m_; A a; a.*foo; // Accesses the m_ member of a }

You can use the somewhat unusual syntax &this->A::m_ to avoid hitting either of these pitfalls.

Example

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