Virtual functions in C++ - Step-by-Step Execution

Concept Flow - Virtual functions

Create Base class pointer

↓

Assign Derived object to pointer

↓

Call virtual function

↓

Check object's actual type

↓

Call Derived's override if exists

↓

Return result

A base class pointer calls a virtual function, which runs the derived class's version if overridden.

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* ptr;
  ptr = new Derived();
  ptr->show();
  delete ptr;
  return 0;
}

Shows how a base class pointer calls the derived class's overridden virtual function.

Execution Table

Step	Action	Pointer Type	Object Type	Function Called	Output
1	Create Base pointer	Base*	None	None
2	Assign Derived object to pointer	Base*	Derived	None
3	Call show() via pointer	Base*	Derived	Derived::show()	Derived
4	End of execution	-	-	-	-

💡 Function call resolved to Derived::show() because show() is virtual and object is Derived

Variable Tracker

Variable	Start	After Step 2	After Step 3	Final
ptr	null	points to Derived object	points to Derived object	points to Derived object

Key Moments - 2 Insights

Why does ptr->show() call Derived::show() even though ptr is a Base*?

Because show() is declared virtual in Base, the program checks the actual object type (Derived) at runtime and calls Derived's version (see execution_table step 3).

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

If show() was not virtual, Base::show() would be called regardless of the object type, because function calls are resolved by pointer type at compile time.

Visual Quiz - 3 Questions

Look at the execution table, what is the output at step 3?

A No output

B Base

C Derived

D Compilation error

💡 Hint

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

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

A Step 2

B Step 3

C Step 1

D Step 4

💡 Hint

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

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

A Derived::show()

B Base::show()

C No function called

D Compiler error

💡 Hint

Refer to key_moments explanation about virtual vs non-virtual function calls

Concept Snapshot

Virtual functions allow a base class pointer to call derived class methods at runtime.
Declare a function with 'virtual' in base class.
Override it in derived class with same signature.
Calls via base pointer use derived version if overridden.
Enables runtime polymorphism in C++.

Full Transcript

This example shows how virtual functions work in C++. We create a base class pointer and assign it a derived class object. When we call the virtual function show() through the base pointer, the program checks the actual object type at runtime. Because the object is of the derived class, the derived class's show() function runs, printing 'Derived'. This is different from normal functions, which are decided at compile time based on pointer type. Virtual functions enable this dynamic behavior, called runtime polymorphism.