C++多重继承，菱形继承中构造函数的调用顺序

浏览数：211 / 时间：2015年06月08日

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：