C++ Primer Plus полностью

void Star::show() const {...} // won't change invoking object

Here const means const Star * this, where this points to the invoking object.

Normally, a function that returns a reference can be on the left side of an assignment statement, which really means you can assign a value to the object referred to. But you can use const to ensure that a reference or pointer return value can’t be used to modify data in an object:

const Stock & Stock::topval(const Stock & s) const

{

    if (s.total_val > total_val)

       return s;           // argument object

    else

       return *this;       // invoking object

}

Here the method returns a reference either to this or to s. Because this and s are both declared const, the function is not allowed to change them, which means the returned reference also must be declared const.

Note that if a function declares an argument as a reference or pointer to a const, it cannot pass along that argument to another function unless that function also guarantees not to change the argument.

Public Inheritance Considerations

Naturally, adding inheritance to a program brings up a number of considerations. Let’s look at a few.

Is-a Relationship Considerations

You should be guided by the is-a relationship. If your proposed derived class is not a particular kind of the base class, you shouldn’t use public derivation. For example, you shouldn’t derive a Programmer class from a Brain class. If you want to represent the belief that a programmer has a brain, you should use a Brain class object as a member of the Programmer class.

In some cases the best approach may be to create an abstract data class with pure virtual functions and to derive other classes from it.

Remember that one expression of the is-a relationship is that a base class pointer can point to a derived-class object and that a base-class reference can refer to a derived-class object without an explicit type cast. Also remember that the reverse is not true; thus, you cannot have a derived-class pointer or reference refer to a base-class object without an explicit type cast. Depending on the class declarations, such an explicit type cast (a downcast) may or may not make sense. (You might want to review Figure 13.4.)

Figure 13.4. Upcasting and downcasting.

What’s Not Inherited

Constructors are not inherited. That is, creating a derived object requires calling a derived-class constructor. However, derived-class constructors typically use the member-initializer list syntax to call on base-class constructors to construct the base class portion of a derived object. If the derived-class constructor doesn’t explicitly call a base-class constructor by using the member-initializer list syntax, it uses the base class’s default constructor. In an inheritance chain, each class can use a member initializer list to pass back information to its immediate base class. C++11 adds a mechanism that enables the inheriting of constructors. However, the default behavior is still that constructors are not inherited.

Destructors are not inherited either. However, when an object is destroyed, the program first calls the derived destructor and then the base destructor. If there is a default base class destructor, the compiler generates a default derived class destructor. Generally speaking, if a class serves as a base class, its destructor should be virtual.

Assignment operators are not inherited. The reason is simple. An inherited method has the same function signature in a derived class as it does in the base class. However, an assignment operator has a function signature that changes from class to class because it has a formal parameter that is the class type. Assignment operators do have some interesting properties, which we’ll look at next.

Assignment Operator Considerations

If the compiler detects that a program assigns one object to another of the same class, it automatically supplies that class with an assignment operator. The default, or implicit, version of this operator uses memberwise assignment, with each member of the target object being assigned the value of the corresponding member of the source object. However, if the object belongs to a derived class, the compiler uses the base-class assignment operator to handle assignment for the base-class portion of the derived-class object. If you’ve explicitly provided an assignment operator for the base class, that operator is used. Similarly, if a class contains a member that is an object of another class, the assignment operator for that class is used for that member.