上一篇说了一般继承,也就是单继承的虚函数表,接下来说说多重继承的虚函数表:
1.无虚函数覆盖的多重继承:
代码:
#pragma once
//无覆盖,多重继承
class Base1
{
public: //三个虚函数
virtual void f() { cout << "Base1::f" << endl; }
virtual void g() { cout << "Base1::g" << endl; }
virtual void h() { cout << "Base1::h" << endl; }
};
class Base2
{
public: //三个虚函数
virtual void f() { cout << "Base2::f" << endl; }
virtual void g() { cout << "Base2::g" << endl; }
virtual void h() { cout << "Base2::h" << endl; }
};
class Base3
{
public: //三个虚函数
virtual void f() { cout << "Base3::f" << endl; }
virtual void g() { cout << "Base3::g" << endl; }
virtual void h() { cout << "Base3::h" << endl; }
};
//多重继承无覆盖
class Derive :public Base1 , public Base2 , public Base3
{
public:
virtual void f1() { cout << "Derive::f1" << endl; }
virtual void g1() { cout << "Derive::g1" << endl; }
virtual void h1() { cout << "Derive::h1" << endl; }
};
void Test()
{
Derive d;
}
调试结果:
可得:
1》每个父类都有虚表;
2》同样问题,虚表中没有体现出子类的虚函数;见真实内容:
可见子类的虚函数在按基类声明顺序的第一个基类的虚表中,且在此基类虚函数之后;
2.有虚函数覆盖的多重继承:
代码:
#pragma once
//有虚函数覆盖的多重继承
#pragma once
//无覆盖,多重继承
class Base1
{
public: //三个虚函数
virtual void f() { cout << "Base1::f" << endl; }
virtual void g() { cout << "Base1::g" << endl; }
virtual void h() { cout << "Base1::h" << endl; }
};
class Base2
{
public: //三个虚函数
virtual void f() { cout << "Base2::f" << endl; }
virtual void g() { cout << "Base2::g" << endl; }
virtual void h() { cout << "Base2::h" << endl; }
};
class Base3
{
public: //三个虚函数
virtual void f() { cout << "Base3::f" << endl; }
virtual void g() { cout << "Base3::g" << endl; }
virtual void h() { cout << "Base3::h" << endl; }
};
//多重继承无覆盖
class Derive :public Base1, public Base2, public Base3
{
public:
virtual void f() { cout << "Derive::f" << endl; } //唯一一个覆盖的子类函数
virtual void g1() { cout << "Derive::g1" << endl; }
virtual void h1() { cout << "Derive::h1" << endl; }
};
void Test()
{
Derive d;
Base1 *b1 = &d;
Base2 *b2 = &d;
Base3 *b3 = &d;
b1->f(); //Derive::f()
b2->f(); //Derive::f()
b3->f(); //Derive::f()
b1->g(); //Base1::g()
b2->g(); //Base2::g()
b3->g(); //Base3::g()
}
运行结果:
其实底层是这样的:
分析:
1》每个父类的虚表中原本存放f()函数的地方,被子类的f()函数覆盖;
2》其余不变;(意思是,还有没被用来覆盖的子类虚函数,任然在首个父类虚函数表的后边位置)
见图:
赐教!
原文链接: https://www.cnblogs.com/li-ning/p/9489966.html
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