C++中多重继承不免会出现钻石继承，也就是继承类的两个基类同时又是同一个基类的继承类，当创建一个对象的时候，他们是按照什么样的顺序调用构造函数的呢。
如果不进行虚拟继承：
class Base { public: Base() { cout<<"Base默认构造函数调用"<<endl; } Base(int i) { cout<<"Base参数构造函数调用"<<endl; cout<<i<<endl; } virtual ~Base(){} }; class Base1: public Base{ public: Base1(int i,int j=0):Base(j){ cout<<"Base1参数构造函数调用"<<endl; cout<<i<<endl; } virtual ~Base1(){} }; class Base2: public Base{ public: Base2(int i):Base(i){ cout<<"Base2参数构造函数调用"<<endl; cout<<i<<endl; } virtual ~Base2(){} }; class Drived:public Base1,public Base2{ public: Drived(int a,int b,int c,int d):Base1(a),Base2(b){ } virtual ~Drived(){} };
新建对象的运行结果：
虚拟继承（虚拟继承中Base1 Base2 中对Base的构造函数调用不再起作用，Base构造函数的调用由derived类直接负责，若Drived不明确指出，则调用默认无参数的构造函数）：
class Base { public: Base() { cout<<"Base默认构造函数调用"<<endl; } Base(int i) { cout<<"Base参数构造函数调用"<<endl; cout<<i<<endl; } virtual ~Base(){} }; class Base1: virtual public Base{ public: Base1(int i,int j=0):Base(j){ cout<<"Base1参数构造函数调用"<<endl; cout<<i<<endl; } virtual ~Base1(){} }; class Base2: virtual public Base{ public: Base2(int i):Base(i){ cout<<"Base2参数构造函数调用"<<endl; cout<<i<<endl; } virtual ~Base2(){} }; class Drived:public Base1,public Base2{ public: Drived(int a,int b,int c,int d):Base1(a),Base2(b){ } virtual ~Drived(){} };
新对象的运行结果：
如果Drived类中有Base1 Base2 类型的成员函数：
class Drived:public Base1,public Base2{ public: Base1 mem1; Base2 mem2; Drived(int a,int b,int c,int d):Base1(a),Base2(b),mem1(c),mem2(d){ } virtual ~Drived(){} };
则新对象的运行结果如图：
如果在Derived类中指定Base的构造函数：
class Drived:public Base1,public Base2{ public: Base1 mem1; Base2 mem2; Drived(int a,int b,int c,int d):Base1(a),Base2(b),Base(a),mem1(c),mem2(d){ } virtual ~Drived(){} };
则运行结果如图所示：
原文：http://www.cnblogs.com/liwenzhu/p/4162705.html