C++学习-派生类的复制运算符

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派生类的赋值控制函数

派生类构造函数在其初始化阶段不仅要初始化类自己的成员，还负责初始化派生类对象的基类部分。因此，派生类的复制控制函数在复制自有成员的同时，也要复制基类部分的成员。类似的，如果派生类的赋值运算符没有处理基类的相应的部分，则派生类中基类的部分会采用默认值，请看下面的程序

#include <iostream>using namespace std;class Base{public:    Base() { p1 = new int; *p1 = 1; cout << "Base::Base()" << endl; }    ~Base() { delete p1; cout << "Base::~Base()" << endl; }    Base& operator=(const Base &b) { cout << "Base::operator=()" << endl; p1 = new int; *p1 = *b.p1; return *this; }    int *p1;};class Derived : public Base{public:    Derived() { p2 = new int; *p2 = 2; cout << "Derived::Derived()" << endl; }    ~Derived() { delete p2; cout << "Derived::~Derived()" << endl; }Derived & operator=(Derived &d) { cout << "Derived::operator=()" << endl; p2 = new int; *p2 = *d.p2; return *this;}    int *p2;};int main(void){    Derived d;    *d.p1 = 3;    *d.p2 = 4;    cout << "d.p1: " << *d.p1 << endl;    cout << "d.p2: " << *d.p2 << endl;//Derived d2 = d; //注意：此时调用的不是赋值运算符，而是d2的复制构造函数！！！Derived d2;d2 = d;    cout << "d2.p1: " << *d2.p1 << endl;    cout << "d2.p2: " << *d2.p2 << endl;    *d2.p1 = 5;    *d2.p2 = 6;    cout << endl;    cout << "d.p1: " << *d.p1 << endl;    cout << "d.p2: " << *d.p2 << endl;    cout << "d2.p1: " << *d2.p1 << endl;    cout << "d2.p2: " << *d2.p2 << endl;    return 0;}

输出结果为：

在没有给d2中的p1和p2重新赋值前，d2的*p1为1而不是为3，显然和我们想要的结果不一致。。。

派生类的赋值运算符也必须地显式为基类部分赋值：

//Base::operator=(const Base&)不会被自动调用Derived &Derived::operator=(const Derived &d){Base::operator=(d); //为基类部分赋值p2 = new int; *p2 = *d.p2;return *this;}

更改后的完整程序：

#include <iostream>using namespace std;class Base{public:    Base() { p1 = new int; *p1 = 1; cout << "Base::Base()" << endl; }    ~Base() { delete p1; cout << "Base::~Base()" << endl; }    Base& operator=(const Base &b) { cout << "Base::operator=()" << endl; p1 = new int; *p1 = *b.p1; return *this; }    int *p1;};class Derived : public Base{public:    Derived() { p2 = new int; *p2 = 2; cout << "Derived::Derived()" << endl; }    ~Derived() { delete p2; cout << "Derived::~Derived()" << endl; }Derived & operator=(Derived &d) { cout << "Derived::operator=()" << endl; p2 = new int; *p2 = *d.p2; return *this;}    int *p2;};int main(void){    Derived d;    *d.p1 = 3;    *d.p2 = 4;    cout << "d.p1: " << *d.p1 << endl;    cout << "d.p2: " << *d.p2 << endl;//Derived d2 = d; //注意：此时调用的不是拷贝控制函数，而是d2的默认复制构造函数Derived d2;d2 = d;    cout << "d2.p1: " << *d2.p1 << endl;    cout << "d2.p2: " << *d2.p2 << endl;    *d2.p1 = 5;    *d2.p2 = 6;    cout << endl;    cout << "d.p1: " << *d.p1 << endl;    cout << "d.p2: " << *d.p2 << endl;    cout << "d2.p1: " << *d2.p1 << endl;    cout << "d2.p2: " << *d2.p2 << endl;    return 0;}

这样就和我们想要的结果一致了

参考：

1、《C++ Prime》第5版 15.7 构造函数与拷贝控制

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