Discovering Inheritance Techniques

If the dontTell() method is called on a Blabber object, it will output I’ll tell all!

myBlabber.dontTell(); // Outputs “I’ll tell all!”

In this case, however, the protected method in the superclass stays protected because any attempts to call Secret’s dontTell() method through a pointer or reference will not compile.

Blabber myBlabber;

Secret& ref = myBlabber;

Secret* ptr = &myBlabber;

ref.dontTell(); // BUG! Attempt to access protected method. ptr->dontTell(); // BUG! Attempt to access protected method.

The only truly useful way to change a method’s access level is by providing a less restrictive accessor to a protected method.

Copy Constructors and the Equals Operator

In Chapter 9, we said that providing a copy constructor and assignment operator is considered a good programming practice when you have dynamically allocated memory in the class. When defining a subclass, you need to be careful about copy constructors and operator=.

If your subclass does not have any special data (pointers, usually) that require a nondefault copy constructor or operator=, you don’t need to have one, regardless of whether or not the superclass has one. If your subclass omits the copy constructor, the parent copy constructor will still be called when the object is copied. Similarly, if you don’t provide an explicit operator=, the default one will be used, and operator= will still be called on the parent class.

On the other hand, if you do specify a copy constructor in the subclass, you need to explicitly chain to the parent copy constructor, as shown in the following code. If you do not do this, the default constructor (not the copy constructor!) will be used for the parent portion of the object.

class Super

{

public:

Super();

Super(const Super& inSuper);

};

class Sub : public Super

{

public:

Sub();

Sub(const Sub& inSub);

};

Sub::Sub(const Sub& inSub) : Super(inSub)

{

}

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Similarly, if the subclass overrides operator=, it is almost always necessary to call the parent’s version of operator= as well. The only case where you wouldn’t do this is if there is some bizarre reason why you only want part of the object assigned when an assignment takes place. The following code shows how to call the parent’s assignment operator from the subclass:

Sub& Sub::operator=(const Sub& inSub)

{

if (&inSub == this) { return *this;

}

Super::operator=(inSub) // Call parent’s operator=.

// Do necessary assignments for subclass.

return (*this);

}

If your subclass does not specify its own copy constructor or operator=, the parent functionality continues to work. If the subclass does provide its own copy constructor or operator=, it needs to explicitly reference the parent versions.

The Truth about Virtual

When you first encountered method overriding above, we told you that only virtual methods can be properly overridden. The reason we had to add the qualifier properly is that if a method is not virtual, you can still attempt to override it but it will be wrong in subtle ways.

Hiding Instead of Overriding

The following code shows a superclass and a subclass, each with a single method. The subclass is attempting to override the method in the superclass, but it is not declared to be virtual in the superclass.

class Super

{

public:

void go() { cout << “go() called on Super” << endl; }

};

class Sub : public Super

{

public:

void go() { cout << “go() called on Sub” << endl; }

};

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Attempting to call the go() method on a Sub object will initially appear to work.

Sub mySub;

mySub.go();

The output of this call is, as expected, go() called on Sub. However, since the method was not virtual, it was not actually overridden. Rather, the Sub class created a new method, also called go() that is completely unrelated to the Super class’s method called go(). To prove this, simply call the method in the context of a Super pointer or reference.

Sub mySub;

Super& ref = mySub;

ref.go();

You would expect the output to be, go() called on Sub, but in fact, the output will be, go() called on Super. This is because the ref variable is a Super reference and because the virtual keyword was omitted. When the go() method is called, it simply executes Super’s go() method. Because it is not virtual, there is no need to consider whether a subclass has overridden it.

Attempting to override a non-virtual method will “hide” the superclass definition and will only be used in the context of the subclass.

How virtual Is Implemented

To understand why method hiding occurs, you need to know a bit more about what the virtual keyword actually does. When a class is compiled in C++, a binary object is created that contains all of the data members and methods for the class. In the non-virtual case, the code to jump to the appropriate method is hard-coded directly where the method is called based on the compile-time type.

If the method is declared virtual, the implementation is looked up in a special area of memory called the vtable, for “virtual table.” Each class that has one or more virtual methods has a vtable that contains pointers to the implementations of the virtual methods. In this way, when a method is called on an object, the pointer is followed into the vtable and the appropriate version of the method is executed based on the type of the object, not the type of the variable used to access it.

Figure 10-11 shows a high-level view of how the vtable makes the overriding of methods possible. The diagram shows two classes, Super and Sub. Super declares two virtual methods, foo() and bar(). As you can see by looking at Super’s vtable, each method has its own implementation defined by the Super class. The Sub class does not override Super’s version of foo(), so the Sub vtable points to the same implementation of foo(). Sub does, however, override bar(), so the vtable points to the new version.

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	Super
	vtable	foo	Super::foo()
			implementation
		bar	Super::bar()
			implementation

Sub

vtable	foo
	bar	Sub::bar()
		implementation

Figure 10-11

The Justification for virtual

Given the fact that you are advised to make all methods virtual, you might be wondering why the virtual keyword even exists. Can’t the compiler automatically make all methods virtual? The answer is yes, it could. Many people think that the language should just make everything virtual. The Java language effectively does this.

The argument against making everything virtual, and the reason that the keyword was created in the first place, has to do with the overhead of the vtable. To call a virtual method, the program needs to perform an extra operation by dereferencing the pointer to the appropriate code to execute. This is a miniscule performance hit for most cases, but the designers of C++ thought that it was better, at least at the time, to let the programmer decide if the performance hit was necessary. If the method was never going to be overridden, there was no need to make it virtual and take the performance hit. There is also a small hit to code size. In addition to the implementation of the method, each object would also need a pointer, which takes up a small, but measurable, amount of space.

The Horror of Non-virtual Destructors

Even programmers who don’t adopt the guideline of making all methods virtual still adhere to the rule when it comes to destructors. The reason is that making your destructors non-virtual can easily cause memory leaks.

For example, if a subclass uses memory that is dynamically allocated in the constructor and deleted in the destructor, it will never be freed if the destructor is never called. As the following code shows, it is easy to “trick” the compiler into skipping the call to the destructor if it is non-virtual.

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