CODEWITHSUNDEEP
c++classinheritance
Sito
Sito

Reputation: 556

Unexpected output of inherited member function

Consider the following code:

#include <list>
#include <iostream>

class test{
    private:
        std::list<double> lst;

    public:
        int get_size(){
            return lst.size();
        }
};

class derived : public test{
    private:
        std::list<double> lst;

    public:
        derived(int u=8){
            for (int k =0 ; k < u; ++k){
                int n = lst.size();
                lst.push_back(0);
            }
        }
};

int main(){

    derived obj = derived();

    std::cout << obj.get_size() << "\n";
    return 0;
}

I'm trying to understand why the output of the above code isn't 8. I thought that the idea of inheritance allows one to use member functions of the base class in the derived class, but it seems that this isn't really working here. Of course I could make the get_size function virtual in the base class and then copy paste the code into the derived class and add override, but this seems really weird, since in this case I could just re-implement the function anyways... So can somebody explain how we need to modify the code from above to get the output 8?

Upvotes: 0

Views: 37

Answers (2)

user10957435
user10957435

Reputation:

You're not using the list from test. You're declaring a new list and using that instead in your constructor for derived.

On the other hand, size() is using the list from test.

If you want to add to the test list from inside derived, you should make test's list protected, not private. And you should remove the list with a duplicate name from derived.

Upvotes: 1

Frank Puffer
Frank Puffer

Reputation: 8215

The lst in your derived class is not the same one as used in your base class.

You should make lst protected instead of private in your test base class. Then you can use the same variable in the derived class. Of course you also need to delete the lst declaration in your derived class.

Upvotes: 1

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