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Function overriding and how to resolve it

Short answer

Function overriding is redefining a base function with identical signature in a derived class. Resolution is static (based on pointer/reference type) for non-virtual functions; dynamic via virtual functions where the actual object type decides at runtime; the overridden base version can be explicitly called with Base::function.

What is function overriding? How can it be resolved? Explain with a suitable example.

Function overriding means a derived class defines a function with the same name, return type and parameters as a function of its base class. The derived version hides the base version for derived objects. The question of "resolving" is deciding which version is called.

Resolution:

  1. Static (compile-time) resolution - the version is chosen by the type of the object or pointer. A derived object calls the derived function; a base pointer calls the base function.
  2. Scope resolution operator - the hidden base version can be called explicitly as Base::fun().
  3. Virtual functions (dynamic resolution) - if the base function is virtual, the version is chosen at run time by the actual object type, even through a base pointer.

Example:

#include <iostream>
using namespace std;

class Base {
public:
    void show()          { cout << "Base::show" << endl; }
    virtual void print() { cout << "Base::print" << endl; }
};

class Derived : public Base {
public:
    void show()  { cout << "Derived::show" << endl; }
    void print() { cout << "Derived::print" << endl; }
};

int main() {
    Derived d;
    Base *p = &d;

    d.show();          // Derived::show (derived object)
    d.Base::show();    // Base::show (scope resolution)
    p->show();         // Base::show (non-virtual: pointer type decides)
    p->print();        // Derived::print (virtual: object type decides)
    return 0;
}

Output:

Derived::show
Base::show
Base::show
Derived::print

✓ Answer checked by Claude Opus 5.5 (high effort)

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