How do you declare an interface in C++?

Question

How do I setup a class that represents an interface? Is this just an abstract base class?

Solution 1

To expand on the answer by bradtgmurray, you may want to make one exception to the pure virtual method list of your interface by adding a virtual destructor. This allows you to pass pointer ownership to another party without exposing the concrete derived class. The destructor doesn't have to do anything, because the interface doesn't have any concrete members. It might seem contradictory to define a function as both virtual and inline, but trust me - it isn't.

class IDemo
{
    public:
        virtual ~IDemo() {}
        virtual void OverrideMe() = 0;
};

class Parent
{
    public:
        virtual ~Parent();
};

class Child : public Parent, public IDemo
{
    public:
        virtual void OverrideMe()
        {
            //do stuff
        }
};

You don't have to include a body for the virtual destructor - it turns out some compilers have trouble optimizing an empty destructor and you're better off using the default.

Solution 2

Make a class with pure virtual methods. Use the interface by creating another class that overrides those virtual methods.

A pure virtual method is a class method that is defined as virtual and assigned to 0.

class IDemo
{
    public:
        virtual ~IDemo() {}
        virtual void OverrideMe() = 0;
};

class Child : public IDemo
{
    public:
        virtual void OverrideMe()
        {
            // do stuff
        }
};

Solution 3

The whole reason you have a special Interface type-category in addition to abstract base classes in C#/Java is because C#/Java do not support multiple inheritance.

C++ supports multiple inheritance, and so a special type isn't needed. An abstract base class with no non-abstract (pure virtual) methods is functionally equivalent to a C#/Java interface.

Solution 4

There is no concept of "interface" per se in C++. AFAIK, interfaces were first introduced in Java to work around the lack of multiple inheritance. This concept has turned out to be quite useful, and the same effect can be achieved in C++ by using an abstract base class.

An abstract base class is a class in which at least one member function (method in Java lingo) is a pure virtual function declared using the following syntax:

class A
{
  virtual void foo() = 0;
};

An abstract base class cannot be instantiated, i. e. you cannot declare an object of class A. You can only derive classes from A, but any derived class that does not provide an implementation of foo() will also be abstract. In order to stop being abstract, a derived class must provide implementations for all pure virtual functions it inherits.

Note that an abstract base class can be more than an interface, because it can contain data members and member functions that are not pure virtual. An equivalent of an interface would be an abstract base class without any data with only pure virtual functions.

And, as Mark Ransom pointed out, an abstract base class should provide a virtual destructor, just like any base class, for that matter.

Solution 5

As far I could test, it is very important to add the virtual destructor. I'm using objects created with new and destroyed with delete.

If you do not add the virtual destructor in the interface, then the destructor of the inherited class is not called.

class IBase {
public:
    virtual ~IBase() {}; // destructor, use it to call destructor of the inherit classes
    virtual void Describe() = 0; // pure virtual method
};

class Tester : public IBase {
public:
    Tester(std::string name);
    virtual ~Tester();
    virtual void Describe();
private:
    std::string privatename;
};

Tester::Tester(std::string name) {
    std::cout << "Tester constructor" << std::endl;
    this->privatename = name;
}

Tester::~Tester() {
    std::cout << "Tester destructor" << std::endl;
}

void Tester::Describe() {
    std::cout << "I'm Tester [" << this->privatename << "]" << std::endl;
}


void descriptor(IBase * obj) {
    obj->Describe();
}

int main(int argc, char** argv) {

    std::cout << std::endl << "Tester Testing..." << std::endl;
    Tester * obj1 = new Tester("Declared with Tester");
    descriptor(obj1);
    delete obj1;

    std::cout << std::endl << "IBase Testing..." << std::endl;
    IBase * obj2 = new Tester("Declared with IBase");
    descriptor(obj2);
    delete obj2;

    // this is a bad usage of the object since it is created with "new" but there are no "delete"
    std::cout << std::endl << "Tester not defined..." << std::endl;
    descriptor(new Tester("Not defined"));


    return 0;
}

If you run the previous code without virtual ~IBase() {};, you will see that the destructor Tester::~Tester() is never called.

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