Access to derived class members through a base class reference

Question

considere a simle class stDeriv which inherits from stBase. I'am surprised to see that we cannot access to the stDeriv class members through a stBase reference.

Below the basic example to illustrate my point :

#include <fstream> // std::ifstream #include <iostream> // std::cout using namespace std; typedef struct stBase { public: virtual ~stBase() { ; } stBase(int iB) { iB_ = iB; } // Copy operator stBase &operator=(const stBase &src) { iB_ = src.iB_; return *this; } virtual void Hello() { cout << " Hello from stBase" << endl; } private: int iB_ = 0; } stBase; typedef struct stDeriv : public stBase { public: int iD_ = 0; stDeriv(int iB) : stBase(iB), iD_(0) { ; } virtual void Hello() { cout << " Hello from stDeriv" << endl; } // Copy operator stDeriv &operator=(const stDeriv &src) { iD_ = src.iD_; return *this; } } stDeriv; int main(int, char *[]) { int iErr = 0; stBase aBase(0); stDeriv aDeriv(1); stDeriv &rDeriv = aDeriv; stBase &rBase = aBase; rBase.Hello(); // OK result : "Hello from stBase" rDeriv.Hello(); // OK result : "Hello from stDeriv" rBase = rDeriv; // KO!!! Cannot access to aDeriv.iD_ through rBase !!! rBase.Hello(); // KO!!! result : "Hello from stBase" !!! return iErr; }

Why do I fail to access to stDeriv::iD_ through rBase after "rBase = rDeriv;" ?

A Deriv is a Base, but a Base is not a Deriv. This is the reason. — Fareanor
– Fareanor, Commented Oct 18, 2021 at 11:00
The typedef struct A {} A; construct is a C-ism that, while formally valid, has no use in C++. — molbdnilo
– molbdnilo, Commented Oct 18, 2021 at 11:49
rBase is a reference to stBase. Look at the definition of the stBase struct. Does it have an iD_ member? No, it doesn't, and that's why you can't access it. — molbdnilo
– molbdnilo, Commented Oct 18, 2021 at 11:52
rBase = rDeriv; does not change the reference, but the referred-to object: in your code it's equivalent to aBase = aDeriv;. — chi
– chi, Commented Oct 18, 2021 at 12:09
After the line "rBase = rDeriv;", I expected to be able to visit aDeriv attributes, which is unfortunately wrong. All is clearly explained in the Pepijn Kramer' s answer below. — sylwa06
– sylwa06, Commented Oct 18, 2021 at 14:30

Pepijn Kramer · Accepted Answer · 2021-10-18 11:00:05Z

You cannot rebind references like you do. rBase already has a value and cannot be assigned to again. why doesn't C++ allow rebinding a reference?

So just make a new reference:

int main(int, char* []) { int iErr = 0; stBase aBase(0); stDeriv aDeriv(1); stDeriv& rDeriv = aDeriv; stBase& rBase = aBase; rBase.Hello(); // OK result : "Hello from stBase" rDeriv.Hello(); // OK result : "Hello from stDeriv" // Make a new reference and all is fine stBase& rBase2 = rDeriv; rBase2.Hello(); // OK result : "Hello from stDeriv" return iErr; }

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