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Inheritance question

General and Gameplay Programming Programming

Started by george7378 February 25, 2014 05:36 PM

10 comments, last by BitMaster 10 years, 1 month ago

george7378

1,443

Author

February 25, 2014 05:36 PM

Hi!

I'm writing a program with a base class and multiple derived classes which are stored in a vector. I think it's easiest if I just post a barebones version to demonstrate my problem:


#include <iostream>

#include <vector>



using namespace std;



class Base

{

public:



    Base(){};



    virtual void identify() const

    {

        cout << "Base" << endl;

    }

};



class Derived1 : public Base

{

public:



    Derived1(){};



    void identify() const

    {

        cout << "Derived1" << endl;

    }

};



class Derived2 : public Base

{

public:



    Derived2(){};



    void identify() const

    {

        cout << "Derived2" << endl;

    }

};



int main()

{

    //Static

    vector <Base> vec;

    vec.push_back(Derived1());

    vec.push_back(Derived2());



    vec[0].identify();

    vec[1].identify();



    //Dynamic

    vector <Base *> vecdyn;

    vecdyn.push_back(new Derived1());

    vecdyn.push_back(new Derived2());



    vecdyn[0]->identify();

    vecdyn[1]->identify();



    while (!vecdyn.empty())

    {

        Base *element = vecdyn.back();

        vecdyn.pop_back();

        delete element;

    }



    cin.get();

    return 0;

}

Running the above program gives me this result:

Base

Derived1

Derived2

...so the Derived1 and Derived2 objects stored in the static vector identify themselves as Base objects while the ones stored in the dynamic vector identify themselves as their respective derive classes. Why are they doing this, and is there a way to make the static objects identify themselves as derived classes rather than base ones?

Much appreciated

Sourceforge projects

BitMaster

8,652

February 25, 2014 05:52 PM

The problem is well known as slicing and there is no way to avoid that without storing pointers instead of instances.

However, modern C++ has plenty of tools to avoid the memory management pitfalls of that (std::unique_ptr/std::shared_ptr among others).

alvaro

21,604

February 25, 2014 05:59 PM

What BitMaster said. I would use std::vector<std::unique_ptr<Base>>.

Also, your base classes should always have a virtual destructor.

frob

46,221

February 25, 2014 05:59 PM

A longer description of the problems.

When you have a base object rather than a base pointer, you are telling the compiler to make certain assumptions. When you use the type directly the language assumes you are not invoking certain behaviors, that the object really is the type you specified and not a derived type.

Also note that you don't have a virtual destructor. A destructor should either be public and virtual (meaning you can use it as a base class) or protected and non-virtual (meaning you cannot inherit from it.) This has the same basic problem, if you don't do the destructor properly the object can be incompletely destroyed.

ChaosEngine

5,385

February 25, 2014 09:03 PM

Also note that you don't have a virtual destructor. A destructor should either be public and virtual (meaning you can use it as a base class) or protected and non-virtual (meaning you cannot inherit from it.) This has the same basic problem, if you don't do the destructor properly the object can be incompletely destroyed.

I know this is the correct thing to do, and I am in no way advocating NOT making it virtual, but in this case does it actually matter in practice?

i.e. Derived has no extra memory overhead, so the allocated space for Base is released? Just a question.

if you think programming is like sex, you probably haven't done much of either.-------------- - capn_midnight

alvaro

21,604

February 26, 2014 01:31 AM

Also note that you don't have a virtual destructor. A destructor should either be public and virtual (meaning you can use it as a base class) or protected and non-virtual (meaning you cannot inherit from it.) This has the same basic problem, if you don't do the destructor properly the object can be incompletely destroyed.

I know this is the correct thing to do, and I am in no way advocating NOT making it virtual, but in this case does it actually matter in practice?
i.e. Derived has no extra memory overhead, so the allocated space for Base is released? Just a question.

The fact that Derived has nothing extra to do in the destructor is a detail that Base shouldn't be aware of. In particular, if that changes in the future, only the code in Derived should have to change.

ChaosEngine

5,385

February 26, 2014 01:49 AM

I know this is the correct thing to do, and I am in no way advocating NOT making it virtual, but in this case does it actually matter in practice?
i.e. Derived has no extra memory overhead, so the allocated space for Base is released? Just a question.

The fact that Derived has nothing extra to do in the destructor is a detail that Base shouldn't be aware of. In particular, if that changes in the future, only the code in Derived should have to change.

Yep, that's why I said making it virtual was the Right Thing To Do(tm).

My question was whether, in this particular instance, a non-virtual destructor would break the code, i.e. is the behaviour undefined?
It was an academic question.

if you think programming is like sex, you probably haven't done much of either.-------------- - capn_midnight

alvaro

21,604

February 26, 2014 02:59 AM

I believe destructing a Derived through a pointer to Base is indeed undefined behavior if Base doesn't have a virtual destructor.

To make matters less clear

, this does work:

#include <iostream>
#include <memory>

struct Base {
  Base() {
    std::cout << "Base\n";
  }

  ~Base() {
    std::cout << "~Base\n";
  }
};

struct Derived : Base {
  Derived() {
    std::cout << "Derived\n";
  }

  ~Derived() {
    std::cout << "~Derived\n";
  }
};

int main() {
  std::shared_ptr<Base> p(new Derived);
  // The constructor of std::shared_ptr<Base> is a template and it will remember to call the correct destructor for the type passed!
}

frob

46,221

February 26, 2014 03:20 AM

If the incorrect function is never called, yes, it is well-defined.

However, rather than just hoping it is correct and that nobody ever misuses it, or through documentation and verification and luck never doing an incorrect function call, it is just so much easier to follow the well-established convention and remove all doubt.