本文小编为大家详细介绍“C++继承中的对象构造与析构和赋值重载实例分析”，内容详细，步骤清晰，细节处理妥当，希望这篇“C++继承中的对象构造与析构和赋值重载实例分析”文章能帮助大家解决疑惑，下面跟着小编的思路慢慢深入，一起来学习新知识吧。
一、构造/析构顺序及继承性
class A{private:int _a;public:A(int a = 0): _a(a){cout << "A()" << this << endl;}~A(){cout << "~A()"<< this <<endl;}};class B : public A{private:int _b;public:B(int b): _b(b), A(){cout << "B()" << this << endl;}~B(){cout << "~B()"<< this <<endl;}};
结论：
构造顺序：先构造基类，后构造派生类
析构顺序：先析构派生类，后析构基类
二、拷贝构造的继承性
class A{private:int _a;public:A(int a = 0): _a(a){cout << "A()" << this << endl;}A(const A& src): _a(src._a){cout << "A(const A& src)"<< this << endl;}~A(){cout << "~A()"<< this <<endl;}};class B : public A{private:int _b;public:B(int b): _b(b), A(){cout << "B()" << this << endl;}B(const B& src): _b(src._b){cout << "B(const B& src)" << this << endl;}~B(){cout << "~B()"<< this <<endl;}};
结论：
先调用基类缺省的构造函数，后调用派生类的拷贝构造函数
若派生类没有缺省构造函数A()，就会报错
疑惑：如何去调用基类的拷贝构造而不是缺省构造
#include<iostream>using namespace std;class A{private:int _a;public:A(int a = 0) : _a(a){cout << "A()" << this << endl;}A(const A& src) : _a(src._a){cout << "A(const A& src)" << this << endl;}~A(){cout << "~A()" << this << endl;}};class B : public A{private:int _b;public:B(int b) : _b(b), A(){cout << "B()" << this << endl;}B(const B& src) : _b(src._b), A(src)//发生赋值兼容规则（切片){cout << "B(const B& src)" << this << endl;}~B(){cout << "~B()" << this << endl;}};int main(){B b(10);B b1(b);return 0;}
结果：
将B类型src传递给A类型的A(const A& src)拷贝构造函数，发生了赋值兼容规则（切片现象）
三、赋值重载不具有继承性
#include<iostream>using namespace std;class A{private:int _a;public:A(int a = 0) : _a(a){cout << "A()" << this << endl;}A(const A& src) : _a(src._a){cout << "A(const A& src)" << this << endl;}A& operator=(const A& src){if(this != &src){_a = src._a;cout << "A& operator=(const A& src)" << endl;}}~A(){cout << "~A()" << this << endl;}};class B : public A{private:int _b;public:B(int b) : _b(b), A(){cout << "B()" << this << endl;}B(const B& src) : _b(src._b), A(src)//发生赋值兼容规则（切片){cout << "B(const B& src)" << this << endl;}B& operator=(const B& src){if(this != &src){_b = src._b;cout << "B& operator=(const B& src)" << endl;}}~B(){cout << "~B()" << this << endl;}};int main(){B b1(10);B b2(20);b1 = b2;return 0;}
结论：默认情况下仅仅调用了派生类的对象的赋值重载，并未调用基类的赋值重载。
解决方案：
#include<iostream>using namespace std;class A{private:int _a;public:A(int a = 0) : _a(a){cout << "A()" << this << endl;}A(const A& src) : _a(src._a){cout << "A(const A& src)" << this << endl;}A& operator=(const A& src){if(this != &src){_a = src._a;cout << "A& operator=(const A& src)" << endl;}}~A(){cout << "~A()" << this << endl;}};class B : public A{private:int _b;public:B(int b) : _b(b), A(){cout << "B()" << this << endl;}B(const B& src) : _b(src._b), A(src)//发生赋值兼容规则（切片){cout << "B(const B& src)" << this << endl;}B& operator=(const B& src){if(this != &src){*(A*)this = src;//将调用基类赋值重载_b = src._b;cout << "B& operator=(const B& src)" << endl;}}~B(){cout << "~B()" << this << endl;}};int main(){B b1(10);B b2(20);b1 = b2;return 0;}
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