C++多重繼承，菱形繼承中構造函數的調用順序

C++中多重繼承不免會出現鑽石繼承，也就是繼承類的兩個基類同時又是同一個基類的繼承類，當創建一個對象的時候，他們是按照什么樣的順序調用構造函數的呢。

如果不進行虛擬繼承：

class Base
{
public:
    Base()
    {
        cout<<"Base默認構造函數調用"<<endl;
    }
    Base(int i)
    {
        cout<<"Base參數構造函數調用"<<endl;
        cout<<i<<endl;
    }
    virtual ~Base(){}
};

class Base1: public Base{
public:
    Base1(int i,int j=0):Base(j){
        cout<<"Base1參數構造函數調用"<<endl;
        cout<<i<<endl;
    }
    virtual ~Base1(){}
};

class Base2: public Base{
public:
    Base2(int i):Base(i){
        cout<<"Base2參數構造函數調用"<<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(){}
};

 

新建對象的運行結果：

虛擬繼承（虛擬繼承中Base1 Base2 中對Base的構造函數調用不再起作用，Base構造函數的調用由derived類直接負責，若Drived不明確指出，則調用默認無參數的構造函數）：

class Base
{
public:
    Base()
    {
        cout<<"Base默認構造函數調用"<<endl;
    }
    Base(int i)
    {
        cout<<"Base參數構造函數調用"<<endl;
        cout<<i<<endl;
    }
    virtual ~Base(){}
};

class Base1: virtual public Base{
public:
    Base1(int i,int j=0):Base(j){
        cout<<"Base1參數構造函數調用"<<endl;
        cout<<i<<endl;
    }
    virtual ~Base1(){}
};

class Base2: virtual public Base{
public:
    Base2(int i):Base(i){
        cout<<"Base2參數構造函數調用"<<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(){}
};

新對象的運行結果：

如果Drived類中有Base1 Base2 類型的成員函數：

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(){}
};

則新對象的運行結果如圖：

 

如果在Derived類中指定Base的構造函數：

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(){}
};

則運行結果如圖所示：