How to call a parent class function from derived class function?

Question

How do I call the parent function from a derived class using C++? For example, I have a class called parent, and a class called child which is derived from parent. Within each class there is a print function. In the definition of the child's print function I would like to make a call to the parents print function. How would I go about doing this?

Answer 1

You can convert this to a pointer of your base class and then call the function from that pointer:

((Base*)this)->foo();

Here is a complete example:

#include <iostream>

class Base
{
public:
    void print()
    {
        std::cout << "Base\n";
    }
};

class Derived : public Base
{
public:
    void print()
    {
        std::cout << "Derived\n";
        ((Base*)this)->print();
    }
};

int main() {
    Derived derived;
    derived.print();
}

Answer 2

In C++, when you have a base class (parent) and a derived class (child), and both classes have a function with the same name, you can call the base class's function from the derived class using the scope resolution operator ::.

Let's say you have a parent class with a print function:

    class Parent {
public:
    void print() {
        // Parent's print functionality
    }
};

And you have a derived class that inherits from the parent:

    class Child : public Parent {
public:
    void print() {
        // Child's print functionality

        // To call the parent's print function:
        Parent::print();
    }
};

In the Child class, within the definition of its print function, you can use Parent::print(); to explicitly call the print function of the parent class. This way, you have control over whether to include the parent's functionality in addition to the child's or to override it completely.

Answer 3

Given a parent class named Parent and a child class named Child, you can do something like this:

class Parent {
public:
    virtual void print(int x);
};

class Child : public Parent {
    void print(int x) override;
};

void Parent::print(int x) {
    // some default behavior
}

void Child::print(int x) {
    // use Parent's print method; implicitly passes 'this' to Parent::print
    Parent::print(x);
}

Note that Parent is the class's actual name and not a keyword.

Answer 4

Call the parent method with the parent scope resolution operator.

Parent::method()

class Primate {
public:
    void whatAmI(){
        cout << "I am of Primate order";
    }
};

class Human : public Primate{
public:
    void whatAmI(){
        cout << "I am of Human species";
    }
    void whatIsMyOrder(){
        Primate::whatAmI(); // <-- SCOPE RESOLUTION OPERATOR
    }
};

Answer 5

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();

        // and when foo() is not called for 'this':
        bottom b;
        b.left::foo();  // calls b.foo() from 'left'
        b.right::foo();  // call b.foo() from 'right'
    }
};

Incidentally, you can't derive directly from the same class twice since there will be no way to refer to one of the base classes over the other.

class bottom : public left, public left { // Illegal
};

Answer 6

If access modifier of base class member function is protected OR public, you can do call member function of base class from derived class. Call to the base class non-virtual and virtual member function from derived member function can be made. Please refer the program.

#include<iostream>
using namespace std;

class Parent
{
  protected:
    virtual void fun(int i)
    {
      cout<<"Parent::fun functionality write here"<<endl;
    }
    void fun1(int i)
    {
      cout<<"Parent::fun1 functionality write here"<<endl;
    }
    void fun2()
    {

      cout<<"Parent::fun3 functionality write here"<<endl;
    }

};

class Child:public Parent
{
  public:
    virtual void fun(int i)
    {
      cout<<"Child::fun partial functionality write here"<<endl;
      Parent::fun(++i);
      Parent::fun2();
    }
    void fun1(int i)
    {
      cout<<"Child::fun1 partial functionality write here"<<endl;
      Parent::fun1(++i);
    }

};
int main()
{
   Child d1;
   d1.fun(1);
   d1.fun1(2);
   return 0;
}

Output:

$ g++ base_function_call_from_derived.cpp
$ ./a.out 
Child::fun partial functionality write here
Parent::fun functionality write here
Parent::fun3 functionality write here
Child::fun1 partial functionality write here
Parent::fun1 functionality write here

Answer 7

If your base class is called Base, and your function is called FooBar() you can call it directly using Base::FooBar()

void Base::FooBar()
{
   printf("in Base\n");
}

void ChildOfBase::FooBar()
{
  Base::FooBar();
}

Answer 8

struct a{
 int x;

 struct son{
  a* _parent;
  void test(){
   _parent->x=1; //success
  }
 }_son;

 }_a;

int main(){
 _a._son._parent=&_a;
 _a._son.test();
}

Reference example.

Answer 9

In MSVC there is a Microsoft specific keyword for that: __super

---

MSDN: Allows you to explicitly state that you are calling a base-class implementation for a function that you are overriding.

// deriv_super.cpp
// compile with: /c
struct B1 {
   void mf(int) {}
};

struct B2 {
   void mf(short) {}

   void mf(char) {}
};

struct D : B1, B2 {
   void mf(short) {
      __super::mf(1);   // Calls B1::mf(int)
      __super::mf('s');   // Calls B2::mf(char)
   }
};

---