How do you call base class method using derived class object?

Answer 1

If the base class method is non-private, the derived class can call the base class method implicitly. However, if the derived class overrides or overloads the method, the base class method must be called explicitly. The following demonstrates explicit calls to base class methods:

#include <iostream>

using namespace std;

class Base

{

public:

Base(){}

void Foo(){ cout << "Base::Foo" << endl; }

};

class Derived : public Base

{

public:

Derived(){}

void Foo(){ cout << "Derived::Foo" << endl; Base::Foo(); }

};

int main()

{

Derived derived;

derived.Foo();

derived.Base::Foo(); // Explicit call to base class method.

return(0);

}

Output:

Derived::Foo

Base::Foo

Base::Foo

Answer 2

I'll take the risk of stating the obvious, you call the function, if it's defined in the base class it's automatically available in the derived class (unless it's private).

If there is a function with the same signature in the derived class you can disambiguate it by adding the base class's name followed by two colons base_class::foo(...). You should note that unlike Java and C#, C++ does not have a keyword for "the base class" (super or base) since C++ supports multiple inheritance which may lead to ambiguity.

class left {

public:

void foo();

};

class right {

public:

void foo();

};

class bottom : public left, public right {

public:

void foo()

{

//base::foo();// ambiguous

left::foo();

right::foo();

}

};