成员函数的重载、覆盖与隐藏
成员函数的重载、覆盖(override)与隐藏很容易混淆,C++程序员必须要搞清楚概念,否则错误将防不胜防。8.2.1 重载与覆盖成员函数被重载的特征:(1)相同的范围(在同一个类中);(2)函数名字相同;(3)参数不同;(4)virtual 关键字可有可无。覆盖是指派生类函数覆盖基类函数,特征是:(1)不同的范围(分别位于派生类与基类);(2)函数名字相同;(3)参数相同;(4)基类函数必须有virtual 关键字。示例8-2-1 中,函数Base::f(int)与Base::f(float)相互重载,而Base::g(void)被Derived::g(void)覆盖。#include <iostream.h>class Base{ public:void f(int x){ cout << "Base::f(int) " << x << endl; }void f(float x){ cout << "Base::f(float) " << x << endl; }virtual void g(void){ cout << "Base::g(void)" << endl;}};class Derived : public Base{ public:virtual void g(void){ cout << "Derived::g(void)" << endl;}};void main(void){ Derived d;Base *pb = &d;pb->f(42); // Base::f(int) 42pb->f(3.14f); // Base::f(float) 3.14
pb->g(); // Derived::g(void)}示例8-2-1 成员函数的重载和覆盖8.2.2 令人迷惑的隐藏规则本来仅仅区别重载与覆盖并不算困难,但是C++的隐藏规则使问题复杂性陡然增加。这里“隐藏”是指派生类的函数屏蔽了与其同名的基类函数,规则如下:(1)如果派生类的函数与基类的函数同名,但是参数不同。此时,不论有无virtual关键字,基类的函数将被隐藏(注意别与重载混淆)。(2)如果派生类的函数与基类的函数同名,并且参数也相同,但是基类函数没有virtual关键字。此时,基类的函数被隐藏(注意别与覆盖混淆)。示例程序8-2-2(a)中:(1)函数Derived::f(float)覆盖了Base::f(float)。(2)函数Derived::g(int)隐藏了Base::g(float),而不是重载。(3)函数Derived::h(float)隐藏了Base::h(float),而不是覆盖。#include <iostream.h>class Base{ public:virtual void f(float x){ cout << "Base::f(float) " << x << endl; }void g(float x){ cout << "Base::g(float) " << x << endl; }void h(float x){ cout << "Base::h(float) " << x << endl; }};class Derived : public Base{ public:virtual void f(float x){ cout << "Derived::f(float) " << x << endl; }void g(int x){ cout << "Derived::g(int) " << x << endl; }void h(float x){ cout << "Derived::h(float) " << x << endl; }};示例8-2-2(a)成员函数的重载、覆盖和隐藏据作者考察,很多C++程序员没有意识到有“隐藏”这回事。由于认识不够深刻,“隐藏”的发生可谓神出鬼没,常常产生令人迷惑的结果。示例8-2-2(b)中,bp 和dp 指向同一地址,按理说运行结果应该是相同的,可事实并非这样。void main(void)
{ Derived d;Base *pb = &d;Derived *pd = &d;// Good : behavior depends solely on type of the objectpb->f(3.14f); // Derived::f(float) 3.14pd->f(3.14f); // Derived::f(float) 3.14// Bad : behavior depends on type of the pointerpb->g(3.14f); // Base::g(float) 3.14pd->g(3.14f); // Derived::g(int) 3 (surprise!)// Bad : behavior depends on type of the pointerpb->h(3.14f); // Base::h(float) 3.14 (surprise!)pd->h(3.14f); // Derived::h(float) 3.14}示例8-2-2(b) 重载、覆盖和隐藏的比较8.2.3 摆脱隐藏隐藏规则引起了不少麻烦。示例8-2-3 程序中,语句pd->f(10)的本意是想调用函数Base::f(int),但是Base::f(int)不幸被Derived::f(char *)隐藏了。由于数字10不能被隐式地转化为字符串,所以在编译时出错。class Base{ public:void f(int x);};class Derived : public Base{ public:void f(char *str);};void Test(void){ Derived *pd = new Derived;pd->f(10); // error}示例8-2-3 由于隐藏而导致错误从示例8-2-3 看来,隐藏规则似乎很愚蠢。但是隐藏规则至少有两个存在的理由: