CODEWITHSUNDEEP
c++derived-class
sunny
sunny

Reputation: 189

Method of base class getting called

I have the following code: 

#include<iostream>
using namespace std;

struct Base{
    void f(int x){
        cout<<"B";
    }
};
struct Derived: public Base {
    virtual void f(double x){
        cout<<"D";
    }
};

int main(){
Derived d;
int x = 5;
d.f(x);
Base *pb = &d;
pb->f(x);

}

It outputs: DB
Even though pb stores the pointer to the derived class. Why is the method of
Base class getting called?

Upvotes: 1

Views: 88

Answers (2)

463035818_is_not_an_ai
463035818_is_not_an_ai

Reputation: 123114

Declaring Derived::f virtual doesn't make Base::f virtual, hence when you call f on a pointer to Base then Base::f is called.

You should declare the method virtual in Base. It will then be also virtual in Derived, you don't need to repeat virtual there. In Derived you should use the override specifier like this:

struct Base{
    virtual void f(int x){
        cout<<"B";
    }
};
struct Derived: public Base {
    void f(double x) override {
        cout<<"D";
    }
};

The override specifier helps to catch mistakes when the method does not actually override an inherited method. For example for the above you will get an error along the line of:

source>:10:10: error: 'void Derived::f(double)' marked 'override', but does not override
   10 |     void f(double x) override {
      |          ^

The parameter types must match when you want to override.

This code prints the expected DD:

#include <iostream>

struct Base {
    virtual void f(int x){
        std::cout << "B";
    }
};
struct Derived: public Base {
    void f(int x) override {
        std::cout << "D";
    }
};

Note that if you don't use override then such mistakes can go unnoticed. When the classes are defined like this:

struct Base{
    virtual void f(int x){
        std::cout<<"B";
    }
};
struct Derived: public Base {
    void f(double x) {
        std::cout<<"D";
    }
};

Then Derived::f does not override Base::f. It only hides it and the output would still be DB.

Live Demo

Upvotes: 2

Muhammad Mujtaba Shafique
Muhammad Mujtaba Shafique

Reputation: 295

It prints D first because of your function call of derived class and then it prints B because you are simply calling the function of base class

Upvotes: -1

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