Effective C++ 笔记 —— Item 32: Make sure public inheritance models "is-a."

Consider this code:

class Rectangle 
{
public:
    virtual void setHeight(int newHeight);
    virtual void setWidth(int newWidth);
    virtual int height() const; // return current values
    virtual int width() const;
    // ...
};

void makeBigger(Rectangle& r) // function to increase r’s area
{
    int oldHeight = r.height();
    r.setWidth(r.width() + 10); // add 10 to r's width
    assert(r.height() == oldHeight); // assert that r's height is unchanged
}

Clearly, the assertion should never fail. makeBigger only changes r's width. Its height is never modified. Now consider this code, which uses public inheritance to allow squares to be treated like rectangles:

class Square: public Rectangle { /*...*/ };
Square s;
// ...

assert(s.width() == s.height()); // this must be true for all squares by inheritance, s is-a Rectangle, so we can increase its area

makeBigger(s); 

assert(s.width() == s.height()); // this must still be true for all squares

 

It's just as clear that this second assertion should also never fail. By definition, the width of a square is the same as its height. But now we have a problem. How can we reconcile the following assertions?

• Before calling makeBigger, s's height is the same as its width;
• Inside makeBigger, s's width is changed, but its height is not;
• After returning from makeBigger, s’s height is again the same as its width. (Note that s is passed to makeBigger by reference, so makeBigger modifies s itself, not a copy of s.)

 

public inheritance asserts that everything that applies to base class objects — everything! — also applies to derived class objects.

 

Things to Remember:

• Public inheritance means ”is-a.“ Everything that applies to base classes must also apply to derived classes, because every derived class object is a base class object.