关于虚继承时派生类调用构造函数问题

[whdugh]

#include <iostream>

using namespace std;
class cbasebase
{
public:
    cbasebase() {cout << "cbasebase()" << endl;}
    cbasebase(const string&) {cout << "cbasebase(const string &)" << endl;}
    virtual ~cbasebase() {cout << "~cbasebase()" << endl;}
};
class cbase1: virtual public cbasebase
{
public:
    cbase1(const string& s): cbasebase(s) {cout << "cbase1(const string&)" << endl;}
    virtual ~cbase1() {cout << "~cbase1()" << endl;}
};
class cbase2:  virtual public cbasebase
{
public:
    cbase2(const string& s): cbasebase(s) {cout << "cbase2(const string&)" << endl;}
    virtual ~cbase2() {cout << "~cbase2()" << endl;}
};
class cbase3
{
public:
    cbase3() {cout << "cbase3()" << endl;}
    virtual ~cbase3() {cout << "~cbase3()" << endl;}
};
class cderived: public cbase1, public cbase2, public cbase3
{
public:
    cderived(const string& s): cbase1(s), cbase2(s) {cout << "cderived(const string&)" << endl;}
    virtual ~cderived() {cout << "~cderived()" << endl;}
};

int main()
{
    cderived a(string("hello world"));  ///cbasebase<> 为何虚继承会调用无参构造函数？
                                        ///cbase1<const string&>
                                        ///cbase2<const string&>
                                        ///cbase3<>
                                        ///cderived<const string&>
                                        ///~cderived<>
                                        ///~cbase3<>
                                        ///~cbase2<>
                                        ///~cbase1<>
                                        ///~cbasebase<>

    return 0;
}

在虚继承时，派生类构造对象时，为什么会调用基类中的无参构造函数啊？

[Universe_Admin]

确认下源代码是否一致

[zhaokai115]

因为derived间接派生自base，调用无参base（）是构造derived自己吧。

[zhaokai115]

嘿嘿，上面说错了。应该是一个diamond问题。derived的结构应该包含下面几方面内容： base // base1和base2共同的部分 base1 // base1多出来的部分 base2 // base2多出来的部分 base3 所以要对它们一一初始化

[金丝龙麟闪电劈]

继承是虚继承，就有如下的继承结构了： cbasebase / \ / \ cbase1 cbase2 cbase3 \ / / \ / / cderived 那么在cderived对象中，只有一个cbasebase子对象。在构造cderived对象时， cbasebase的构造函数也最先执行，与cbase1,cbase2的构造函数中对cbasebase的构造函数的调用无关，因为这是在构造cderived对象。 如果想不调用cbasebase的默认构造函数，可改为： class cderived: public cbase1, public cbase2, public cbase3 { public: cderived(const string& s): cbase1(s), cbase2(s),cbasebase(s)

[金丝龙麟闪电劈]

上例中cbase1, cbase2,的构造函数并不会被调用。 请看《inside C++ model》中文版第210页，有详细介绍