Programming Forums > Application Development > C++

gizzy · Feb 15th, 2006, 4:57 PM

I have been searching many places for information on how to create interface classes with C++. I have read about one way which involves pure virtual functions, but that does not solve my problem. I have been practicing Java for about 1.5 years, and would like to know how to create Java type interface classes with C++. I have been practicing C++ for a few months now. I coded a binary search tree in Java, and I now want to code it in C++. An example of what I want to be able to do:

(Toggle Plain Text)

public Interface Comparator...
...public bool eq(Comparator x);

public class MyCharacter implements Comparator
...public bool eq(Comparator x)
   {
       return (getValue() == x.getValue());
   }...

Basically, in Java I would be able to pass a MyCharacter object into a method as a Comparator object. Any ideas as to how I could go about this in C++?

Thanks.

Arevos · Feb 15th, 2006, 6:12 PM

Quote:

Originally Posted by gizzy

I have been searching many places for information on how to create interface classes with C++. I have read about one way which involves pure virtual functions, but that does not solve my problem.

Why doesn't it?

Unlike Java, C++ has multiple inheritance, so interfaces are not strictly needed. The nearest thing to them is an abstract class, which is any C++ class with pure virtual member functions. A pure virtual function is defined thus:

(Toggle Plain Text)

class Foobar {
    public:
        virtual void pure_virtual_function() = 0;
};

An "interface" would be a class consisting entirely of virtual functions.

gizzy · Feb 15th, 2006, 6:47 PM

Whenever I try to use 'Comparator' in a function, like this:

(Toggle Plain Text)

public bool eq(Comparator x)

I get an error saying that I can't create an instance of an abstract class. I'm just unsure as to how I would create a Comparator interface class, so that more than 1 type of class could implement it, and pass instances of those classes into the functions, such as eq

Thanks.

DaWei · Feb 15th, 2006, 7:48 PM

Don't instantiate it; derive from it and provide the methods. A virtual function can't DO anything.

Cerulean · Feb 15th, 2006, 8:36 PM

Quote:

Originally Posted by DaWei

A virtual function can't DO anything.

Just for the sake of clarification - a pure virtual function, that is. A regular virtual method (using just the virtual keyword) can have an implementation and do stuff.

DaWei · Feb 15th, 2006, 9:15 PM

Correct the missing word in my clarification by extrapolating from Arevos' post

.

InfoGeek · Feb 16th, 2006, 8:26 AM

Quote:

Originally Posted by Dawei

Don't instantiate it; derive from it and provide the methods. A virtual function can't DO anything.

Here's an example:

(Toggle Plain Text)

class animal
{
  public:
     virtual void speak()=0;
};

class dog:public animal
{
   public:
      void speak()
        {
          cout<<"bow bow...";
          }
};

class cat:public animal
{
   public:
        void speak()
          {
            cout<<"meow meow...";
            }
 };

int main()
{
   animal *p;
   p = new dog;
   p->speak();//prints bow bow...
   delete p;
   p = new cat;
   p->speak();//prints meow meow...
   delete p;
   return 0;
  }

Hope that helps.

grumpy · Feb 16th, 2006, 9:39 AM

1) An abstract base class in C++ is a class that declares one or more pure virtual member functions. Optionally, in C++, any virtual member function can be defined whether it is pure or not (i.e. contrary to Dawei's and Cerulean's comments, an abstract base class CAN actually provide a body for any virtual function, whether it is pure or not). The existence of any virtual member functions, however, means that the class cannot be instantiated (just as it is not possible to create an instance of a Java interface).

2) A Java interface is approximately functionally equivalent to an abstract base class in C++, with three additional properties;

a) ALL member functions (other than constructors, which are optional) are pure virtual
b) The class has no data members;
c) If the abstract base clase is derived from another, the base class also has these properties.

3) There is no "implements" keyword in C++. The C++ equivalent of implementing a Java interface is deriving from the abstract base class, declaring all pure virtual function(s), and implementing them. C++ supports multiple inheritence of classes, so the Java notion of implementing multiple interfaces can be achieved by inheriting from multiple abstract base classes. In doing this, if ANY inherited pure virtual functions are not overridded (i.e. implemented by the derived class) the derived class is also an abstract base class and cannot be instantiated. Unlike Java, C++ also supports multiple inheritence of class implementation, not just interface.

4) One other property of an abstract base class is that it cannot be passed by value as an argument to a function. It must be passed by reference. Using InfoGeek's example, we can't do this;

(Toggle Plain Text)

void Talk(animal a)
{
     a.speak();
}

but we can do this;

(Toggle Plain Text)

void Talk(animal &a)
{
     a.speak();
}

The second version of the function can be used like this;

(Toggle Plain Text)

int main()
{
        cat c;
        Talk(c);   // prints meow meow 
        dog d;
        Talk(d);   // prints bow wow
}

The reason passing an abstract base class by value is not allowed is that passing by value implicitly creates a copy of the object. Creating a copy implies creating an instance of the abstract base class. And that is not allowed......

gizzy · Feb 16th, 2006, 6:23 PM

Thanks for the help. I have progressed a bit, but am faced with another problem. My Comparator class is like this...

(Toggle Plain Text)

class Comparator
{
public:
	virtual bool eq(Comparator &x) = 0;
	virtual bool ge(Comparator &x) = 0;
	virtual bool gt(Comparator &x) = 0;
};

I have a class CharCon (a character container class) which inherits from the Comparator class, like so...

(Toggle Plain Text)

class CharCon : public Comparator
{
public:
	CharCon();
	CharCon(char initChar);
	~CharCon();

	void setValue(char newChar);
	char getValue();
	void printValue();

	bool eq(Comparator &x);
	bool ge(Comparator &x);
	bool gt(Comparator &x);

private:
	char character;
};

In my implementation of the eq, ge, and gt functions in the CharCon.cpp file, I need to be able to cast the Comparator argument as a CharCon, but the compiler tells me I cannot convert from Comparator to CharCon. This is the eq method in CharCon.cpp...

(Toggle Plain Text)

bool CharCon::eq(Comparator &x)
{
	return (character == ((CharCon)x).getValue());
}

I try to cast x as a CharCon, but it can't be done, and I don't know how I could go about it in another way. Help appreciated. Thanks again.

Kaja Fumei · Feb 16th, 2006, 6:38 PM

But you can convert a pointer to x to a CharCon* like so:

(Toggle Plain Text)

bool CharCon::eq(Comparator &x)
{
	return (character == ((CharCon*) &x)->getValue());
}

or:

(Toggle Plain Text)

bool CharCon::eq(Comparator *x)
{
	return (character == ((CharCon*) x)->getValue());
}

Feb 15th, 2006, 4:57 PM	#1
gizzy Newbie Join Date: Feb 2006 Location: Glasgow, Scotland Posts: 8 Rep Power: 0	Java-style interfaces in C++ I have been searching many places for information on how to create interface classes with C++. I have read about one way which involves pure virtual functions, but that does not solve my problem. I have been practicing Java for about 1.5 years, and would like to know how to create Java type interface classes with C++. I have been practicing C++ for a few months now. I coded a binary search tree in Java, and I now want to code it in C++. An example of what I want to be able to do: (Toggle Plain Text) public Interface Comparator... ...public bool eq(Comparator x); public class MyCharacter implements Comparator ...public bool eq(Comparator x) { return (getValue() == x.getValue()); }... public Interface Comparator... ...public bool eq(Comparator x); public class MyCharacter implements Comparator ...public bool eq(Comparator x) { return (getValue() == x.getValue()); }... Basically, in Java I would be able to pass a MyCharacter object into a method as a Comparator object. Any ideas as to how I could go about this in C++? Thanks. Last edited by gizzy; Feb 15th, 2006 at 5:07 PM.

Feb 15th, 2006, 6:47 PM	#3
gizzy Newbie Join Date: Feb 2006 Location: Glasgow, Scotland Posts: 8 Rep Power: 0	Whenever I try to use 'Comparator' in a function, like this: (Toggle Plain Text) public bool eq(Comparator x) public bool eq(Comparator x) I get an error saying that I can't create an instance of an abstract class. I'm just unsure as to how I would create a Comparator interface class, so that more than 1 type of class could implement it, and pass instances of those classes into the functions, such as eq Thanks.

Feb 15th, 2006, 7:48 PM	#4
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Don't instantiate it; derive from it and provide the methods. A virtual function can't DO anything. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Feb 15th, 2006, 9:15 PM	#6
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Correct the missing word in my clarification by extrapolating from Arevos' post . __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Feb 16th, 2006, 9:39 AM	#8
grumpy Programming Guru Join Date: Jun 2005 Location: Adelaide, South Australia Posts: 1,261 Rep Power: 5	1) An abstract base class in C++ is a class that declares one or more pure virtual member functions. Optionally, in C++, any virtual member function can be defined whether it is pure or not (i.e. contrary to Dawei's and Cerulean's comments, an abstract base class CAN actually provide a body for any virtual function, whether it is pure or not). The existence of any virtual member functions, however, means that the class cannot be instantiated (just as it is not possible to create an instance of a Java interface). 2) A Java interface is approximately functionally equivalent to an abstract base class in C++, with three additional properties; a) ALL member functions (other than constructors, which are optional) are pure virtual b) The class has no data members; c) If the abstract base clase is derived from another, the base class also has these properties. 3) There is no "implements" keyword in C++. The C++ equivalent of implementing a Java interface is deriving from the abstract base class, declaring all pure virtual function(s), and implementing them. C++ supports multiple inheritence of classes, so the Java notion of implementing multiple interfaces can be achieved by inheriting from multiple abstract base classes. In doing this, if ANY inherited pure virtual functions are not overridded (i.e. implemented by the derived class) the derived class is also an abstract base class and cannot be instantiated. Unlike Java, C++ also supports multiple inheritence of class implementation, not just interface. 4) One other property of an abstract base class is that it cannot be passed by value as an argument to a function. It must be passed by reference. Using InfoGeek's example, we can't do this; (Toggle Plain Text) void Talk(animal a) { a.speak(); } void Talk(animal a) { a.speak(); } but we can do this; (Toggle Plain Text) void Talk(animal &a) { a.speak(); } void Talk(animal &a) { a.speak(); } The second version of the function can be used like this; (Toggle Plain Text) int main() { cat c; Talk(c); // prints meow meow dog d; Talk(d); // prints bow wow } int main() { cat c; Talk(c); // prints meow meow dog d; Talk(d); // prints bow wow } The reason passing an abstract base class by value is not allowed is that passing by value implicitly creates a copy of the object. Creating a copy implies creating an instance of the abstract base class. And that is not allowed......

Feb 16th, 2006, 6:23 PM	#9
gizzy Newbie Join Date: Feb 2006 Location: Glasgow, Scotland Posts: 8 Rep Power: 0	Thanks for the help. I have progressed a bit, but am faced with another problem. My Comparator class is like this... (Toggle Plain Text) class Comparator { public: virtual bool eq(Comparator &x) = 0; virtual bool ge(Comparator &x) = 0; virtual bool gt(Comparator &x) = 0; }; class Comparator { public: virtual bool eq(Comparator &x) = 0; virtual bool ge(Comparator &x) = 0; virtual bool gt(Comparator &x) = 0; }; I have a class CharCon (a character container class) which inherits from the Comparator class, like so... (Toggle Plain Text) class CharCon : public Comparator { public: CharCon(); CharCon(char initChar); ~CharCon(); void setValue(char newChar); char getValue(); void printValue(); bool eq(Comparator &x); bool ge(Comparator &x); bool gt(Comparator &x); private: char character; }; class CharCon : public Comparator { public: CharCon(); CharCon(char initChar); ~CharCon(); void setValue(char newChar); char getValue(); void printValue(); bool eq(Comparator &x); bool ge(Comparator &x); bool gt(Comparator &x); private: char character; }; In my implementation of the eq, ge, and gt functions in the CharCon.cpp file, I need to be able to cast the Comparator argument as a CharCon, but the compiler tells me I cannot convert from Comparator to CharCon. This is the eq method in CharCon.cpp... (Toggle Plain Text) bool CharCon::eq(Comparator &x) { return (character == ((CharCon)x).getValue()); } bool CharCon::eq(Comparator &x) { return (character == ((CharCon)x).getValue()); } I try to cast x as a CharCon, but it can't be done, and I don't know how I could go about it in another way. Help appreciated. Thanks again.

Feb 16th, 2006, 6:38 PM	#10
Kaja Fumei Hobbyist Programmer Join Date: Oct 2005 Posts: 134 Rep Power: 4	But you can convert a pointer to x to a CharCon* like so: (Toggle Plain Text) bool CharCon::eq(Comparator &x) { return (character == ((CharCon) &x)->getValue()); } bool CharCon::eq(Comparator &x) { return (character == ((CharCon) &x)->getValue()); } or: (Toggle Plain Text) bool CharCon::eq(Comparator x) { return (character == ((CharCon) x)->getValue()); } bool CharCon::eq(Comparator x) { return (character == ((CharCon) x)->getValue()); }

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Programming Forums > Application Development > C++

Java-style interfaces in C++