Question

Given a base class A, with the A::get and A::add functions defined.

class A { 
public:
  int get() { return 3; }
  int add() {
    return get() + get();
  }
};

Note that A::add calls A::get. Is there a way a derived class B can call A::add but use it’s own B::get function? Something like:

class B : public A {
  int get() { return 7; }
};

#include <iostream>

int main() {
  A foo;
  std::cout << foo.add() << std::endl;

  B bar;
  std::cout << bar.add() << std::endl;
  return 0;
}

Where the expected output is 14 not 6 in the second case. If had control over A, I could make ::get a virtual function and each derived class could implement it as needed. However, lets assume that A is immutable – how do I call the correct ::get?

Answer 1

No, this is not possible: imagine that A is compiled separately and given to you as a pre-compiled library to see why.

Since A::get is not virtual, the compiler inserts a call straight to A::get, without any indirection through vtable, into its output for A::add. Moreover, the compiler could even inline the call if it chooses to do so, “baking in” the return of 3 into its output.