Runtime polymorphism in C++ - Step-by-Step Execution

Concept Flow - Runtime polymorphism

Create base class pointer

↓

Assign derived class object

↓

Call virtual function

↓

At runtime, decide which function to call

↓

Execute derived class function if overridden

↓

End

Runtime polymorphism lets a program decide which function to call during execution using base class pointers to derived objects.

Execution Sample

C++

#include <iostream>
using namespace std;
class Base {
public:
  virtual void show() { cout << "Base"; }
};
class Derived : public Base {
public:
  void show() override { cout << "Derived"; }
};
int main() {
  Base* b = new Derived();
  b->show();
  delete b;
  return 0;
}

This code creates a base class pointer to a derived class object and calls a virtual function, showing runtime polymorphism.

Execution Table

Step	Action	Pointer Type	Object Type	Function Called	Output
1	Create Derived object	-	Derived	-	-
2	Assign Derived object to Base pointer	Base*	Derived	-	-
3	Call show() via Base pointer	Base*	Derived	Derived::show()	Derived
4	Program ends	-	-	-	-

💡 Program ends after calling Derived::show() via Base pointer due to virtual function.

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	Final
b (Base*)	nullptr	nullptr	points to Derived object	points to Derived object	points to Derived object

Key Moments - 2 Insights

Why does calling show() on a Base pointer call Derived's show()?

Because show() is declared virtual in Base, the program decides at runtime to call Derived's version, as shown in execution_table step 3.

What if show() was not virtual? Which function would be called?

If show() was not virtual, Base's show() would be called even if the pointer points to Derived, because function binding would happen at compile time.

Visual Quiz - 3 Questions

Look at the execution_table, what is the output at step 3?

A Base

B Derived

C No output

D Compilation error

💡 Hint

Check the 'Output' column in execution_table row for step 3.

At which step does the Base pointer start pointing to the Derived object?

A Step 3

B Step 1

C Step 2

D Step 4

💡 Hint

Look at the 'Pointer Type' and 'Object Type' columns in execution_table.

If show() was not virtual, how would the function called at step 3 change?

A Base::show() would be called

B No function would be called

C Derived::show() would still be called

D Program would crash

💡 Hint

Refer to key_moments explanation about virtual keyword effect.

Concept Snapshot

Runtime polymorphism in C++ uses virtual functions.
Base class declares a function virtual.
Derived class overrides it.
Base class pointer can point to Derived object.
Calling virtual function via base pointer calls derived version at runtime.

Full Transcript

Runtime polymorphism allows a program to decide which function to call while running. In C++, this is done using virtual functions. A base class declares a function as virtual. A derived class overrides this function. When a base class pointer points to a derived class object and calls the virtual function, the derived class's version runs. This is shown in the example where a Base pointer points to a Derived object. Calling show() on the Base pointer calls Derived's show() because of the virtual keyword. If the function was not virtual, Base's version would run instead. This behavior is called dynamic dispatch and is key to runtime polymorphism.