c++ vitual继承

为了解决多继承下的二义性而设计的vitrul继承

class Base
{
public:
    Base(void);
    ~Base(void);
};

Base::Base(void)
{
    printf("Base::Base():%p\n",this);
}

class ChildFirst :
    /*vitual*/ public Base
{
public:
    ChildFirst(void);
    ~ChildFirst(void);
};

ChildFirst::ChildFirst(void)
{
    printf("ChildFirst::ChildFirst()\n");
}

class ChildSencond :
    /* vitrual */public Base
{
public:
    ChildSencond(void);
    ~ChildSencond(void);
};


ChildSencond::ChildSencond(void)
{
    printf("ChildSencond::ChildSencond()\n");
}


class GrandChild :public ChildFirst, ChildSencond
{
public:
    GrandChild(void);
    ~GrandChild(void);
};

GrandChild::GrandChild(void)
{
    printf("GrandChild::GrandChild()\n");
}

int _tmain(int argc, _TCHAR* argv[])
{
    GrandChild grandChild;
    return 0;
}

这样的话，Base::()Base()会被调用两次而且两次地址不同。

输出

Base::Base():0x0015ff0f

ChildFirst::ChildFirst()

Base::Base():0x0015ff10

ChildSecond::ChildSecond()

GrandChild::GrandChild()

.为解决上述问题，请把注释了virtual打开，注意写法，vitual是直接继承的那里写的。

打开后的输出为：

Base::Base():0x0015000f

ChildFirst::ChildFirst()

ChildSecond::ChildSecond()

GrandChild::GrandChild()