C++面向对象知识点十一:虚基类

在多重继承关系中,会存在如下关系:

class A
{
protected: 
    int x;
public:
    A(int a = 0): x(a) {}
};

class B: public A
{
protected: 
    int y;
public:
    B(int a = 0, int b = 0): A(a), y(b) {}
};

class C: public A
{
protected: 
    int z;
public:
    C(int a = 0, int c = 0): A(a), z(c) {}
};

class D: public B, public C
{
protected: 
    int k;
public:
    D(int a1 = 0, int a2 = 0, int b = 0, int c = 0, int d = 0): B(a1, b), C(a2, c), k(d) {}

    void show()
    {
        cout << "x1=" << B::x << ',';
        cout << "x2=" << C::x << ',';    
        cout << "y=" << y << ',';
        cout << "z=" << z << ',';
        cout << "k=" << k << endl;
    }

};

int main()
{
	D d(1, 2, 3, 4, 5);
	d.show();
	return 0;
}

在上面的程序中,派生类中对基类A做了两次拷贝,若只想拷贝一次,则需要将基类A说明成虚基类。

class B: virtual public A
{
protected: 
    int y;
public:
    B(int a = 0, int b = 0): A(a), y(b) {}
};

class C: virtual public A
{
protected: 
    int z;
public:
    C(int a = 0, int c = 0): A(a), z(c) {}
};

class D: public B, public C
{
protected: 
    int k;
public:
    D(int a1 = 0, int a2 = 0, int b = 0, int c = 0, int d = 0): B(a1, b), C(a2, c), k(d) {}

    void show()
    {
        cout << "x=" << x << ',';  
        cout << "y=" << y << ',';
        cout << "z=" << z << ',';
        cout << "k=" << k << endl;
    }

};

int main()
{
	D d(1, 2, 3, 4, 5);
	d.show();
	return 0;
}

此时,先调用虚基类的构造函数,在调用非虚基类的构造函数。

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转载自blog.csdn.net/Doutd_y/article/details/82115350