多重继承
- C++支持编写多重继承的代码
- 一个子类可以拥有多个父类
- 子类拥有所有父类的成员变量
- 子类继承所有父类的成员函数
- 子类对象可以当作任意父类对象使用
- 多重继承的语法规则
//多重继承的本质与单继承相同
class Derived : public BaseA,
public BaseB,
public BaseC
{
// ......
};
多重继承的问题一
这里首先看一段代码:
#include <iostream>
#include <string>
using namespace std;
//定义父类BaseA
class BaseA
{
//成员变量
int ma;
public:
//带参构造函数
BaseA(int a)
{
ma = a;
}
//成员函数
int getA()
{
return ma;
}
};
//定义父类BaseB
class BaseB
{
//成员变量
int mb;
public:
//带参构造函数
BaseB(int b)
{
mb = b;
}
//成员函数
int getB()
{
return mb;
}
};
//子类Derived同时继承BaseA 和 BaseB
class Derived : public BaseA, public BaseB
{
//子类的成员变量
int mc;
public:
//子类带参构造函数 初始化列表中初始化2个父类
Derived(int a, int b, int c) : BaseA(a), BaseB(b)
{
mc = c;
}
//成员函数
int getC()
{
return mc;
}
//成员函数
void print()
{
cout << "ma = " << getA() << ", "
<< "mb = " << getB() << ", "
<< "mc = " << mc << endl;
}
};
int main()
{
cout << "sizeof(Derived) = " << sizeof(Derived) << endl; // 12
Derived d(1, 2, 3);
d.print();
cout << "d.getA() = " << d.getA() << endl;
cout << "d.getB() = " << d.getB() << endl;
cout << "d.getC() = " << d.getC() << endl;
cout << endl;
return 0;
}
上述代码是一个非常标准的多重继承的示例,输出结果如下:
sizeof(Derived) = 12
ma = 1, mb = 2, mc = 3
d.getA() = 1
d.getB() = 2
d.getC() = 3
因为子类初始化时在初始化列表中也初始化了2个父类,所以调用父类函数可以输出从父类继承的成员变量。此时,我们再做一点其它操作:
//使用子类对象指针分别赋值给父类指针
BaseA* pa = &d;
BaseB* pb = &d;
//分别调用函数
cout << "pa->getA() = " << pa->getA() << endl;
cout << "pb->getB() = " << pb->getB() << endl;
cout << endl;
//使用2个指针分别指向父类指针
void* paa = pa;
void* pbb = pb;
//判断2个指针是否相等
if( paa == pbb )
{
cout << "Pointer to the same object!" << endl;
}
else
{
cout << "Error" << endl;
}
//再分别输出4个指针
cout << "pa = " << pa << endl;
cout << "pb = " << pb << endl;
cout << "paa = " << paa << endl;
cout << "pbb = " << pbb << endl;
输出结果为:
sizeof(Derived) = 12
ma = 1, mb = 2, mc = 3
d.getA() = 1
d.getB() = 2
d.getC() = 3
pa->getA() = 1
pb->getB() = 2
Error
pa = 0x7fff58957a00
pb = 0x7fff58957a04
paa = 0x7fff58957a00
pbb = 0x7fff58957a04
有没有发现一个奇怪的事情,对子类对象d取地址后分别赋值给2个父类指针,此时竟然会得到2个不同的地址。也就是说 **通过多重继承得到的对象可能拥有不同的地址**
!如下所示:
Derived d(1, 2, 3);
BaseA* pa = &d;
BaseB* pb = &d;
pa ====> int ma;
int mb; <==== pb
int mc;
多重继承的问题二
接下来再看一段代码:
#include <iostream>
#include <string>
using namespace std;
class People
{
string m_name;
int m_age;
public:
People(string name, int age)
{
m_name = name;
m_age = age;
}
void print()
{
cout << "Name = " << m_name << ", "
<< "Age = " << m_age << endl;
}
};
class Teacher : virtual public People
{
public:
Teacher(string name, int age) : People(name, age)
{
}
};
class Student : virtual public People
{
public:
Student(string name, int age) : People(name, age)
{
}
};
class Doctor : public Teacher, public Student
{
public:
Doctor(string name, int age) : Teacher(name, age), Student(name, age), People(name, age)
{
}
};
int main()
{
Doctor d("Delphi", 33);
d.print();
return 0;
}
在上述代码中,Teacher类和Student类都是People类的子类,然后Doctor类又同时继承Student类和Teacher类,这里会引起歧义。先看下输出结果:
Name = Delphi, Age = 33
看到结果,可以自问一下,d调用print函数,这个print函数是Teacher类的还是Student类的呢?其实这里反映出多重继承的第二个问题: 当多重继承关系出现闭合时将产生数据冗余的问题!
那如何解决呢? 可是使用 虚继承
:
class People{ };
class Teacher : virtual public People{};
class Student : virtual public People{};
class Doctor : public Teacher, public Student
{
};
- 虚继承能够解决数据冗余问题
- 中间层父类不再关系顶层父类的初始化
- 最终子类必须直接调用顶层父类的构造函数
多重继承的问题三
在问题二中,使用虚继承的方式来解决多继承闭合造成的数据冗余问题,但是又引发了另一个问题:
当架构设计中需要继承时,无法确定使用直接继承还是虚继承!!
多重继承的问题四
接下来看第四个问题,先看一段代码:
#include <iostream>
#include <string>
using namespace std;
//定义BaseA类
class BaseA
{
public:
//定义虚函数funcA()
virtual void funcA()
{
cout << "BaseA::funcA()" << endl;
}
};
//定义BaseB类
class BaseB
{
public:
//定义虚函数funcB()
virtual void funcB()
{
cout << "BaseB::funcB()" << endl;
}
};
//定义Derived类 并同时继承BaseA和BaseB类
class Derived : public BaseA, public BaseB
{
};
int main()
{
Derived d;
//用子类对象d分别赋值给2个父类指针
BaseA* pa = &d;
BaseB* pb = &d;
//把BaseA指针直接强转为BaseB指针
BaseB* pbe = (BaseB*)pa;
//使用dynamic_cast来转换指针
BaseB* pbc = dynamic_cast<BaseB*>(pa);
cout << "sizeof(d) = " << sizeof(d) << endl;
cout << "Using pa to call funcA()..." << endl;
pa->funcA();
cout << "Using pb to call funcB()..." << endl;
pb->funcB();
cout << "Using pbe to call funcB()..." << endl;
pbe->funcB();
cout << "Using pbc to call funcB()..." << endl;
pbc->funcB();
cout << endl;
cout << "pa = " << pa << endl;
cout << "pb = " << pb << endl;
cout << "pbe = " << pbe << endl;
cout << "pbc = " << pbc << endl;
return 0;
}
运行结果如下:
sizeof(d) = 16
Using pa to call funcA()...
BaseA::funcA()
Using pb to call funcB()...
BaseB::funcB()
Using pbe to call funcB()...
BaseA::funcA()
Using pbc to call funcB()...
BaseB::funcB()
pa = 0x7fff59f519f8
pb = 0x7fff59f51a00
pbe = 0x7fff59f519f8
pbc = 0x7fff59f51a00
有没有发现在转换指针时,如果直接进行强制转换,调用的还是原指针指向的方法,也就是说pbe指针虽然是父类指针BaseB类型,但是实际是父类指针BaseA类型,所以调用funcB函数时输出funcA。但如果使用dynamic_cast的时候,就不会存在这个问题。
所以在需要进行强制类型转换时,C++中推荐使用新式类型转换关键字: dynamic_cast
。
正确使用多重继承
- 工程开发中的 "多重继承" 方式:
- 单继承某个类 + 实现(多个)接口
就像这样:
#include <iostream>
#include <string>
using namespace std;
//定义父类Base
class Base
{
protected:
//成员变量mi,子类可访问,外界不可访问
int mi;
public:
//带参构造函数
Base(int i)
{
mi = i;
}
//成员函数
int getI()
{
return mi;
}
//这个函数的作用是判断obj是否是当前对象
bool equal(Base* obj)
{
return (this == obj);
}
};
//定义接口Interface1
class Interface1
{
public:
//2个纯虚函数
virtual void add(int i) = 0;
virtual void minus(int i) = 0;
};
//定义接口Interface2
class Interface2
{
public:
//2个纯虚函数
virtual void multiply(int i) = 0;
virtual void divide(int i) = 0;
};
//定义子类 并继承父类Base,并实现2个接口
class Derived : public Base, public Interface1, public Interface2
{
public:
//初始化时 初始化Base类
Derived(int i) : Base(i)
{
}
void add(int i)
{
mi += i;
}
void minus(int i)
{
mi -= i;
}
void multiply(int i)
{
mi *= i;
}
void divide(int i)
{
if( i != 0 )
{
mi /= i;
}
}
};
int main()
{
Derived d(100);
Derived* p = &d;
Interface1* pInt1 = &d;
Interface2* pInt2 = &d;
cout << "p->getI() = " << p->getI() << endl; // 100
pInt1->add(10);
pInt2->divide(11);
pInt1->minus(5);
pInt2->multiply(8);
cout << "p->getI() = " << p->getI() << endl; // 40
cout << endl;
cout << "pInt1 == p : " << p->equal(dynamic_cast<Base*>(pInt1)) << endl;
cout << "pInt2 == p : " << p->equal(dynamic_cast<Base*>(pInt2)) << endl;
return 0;
}
输出结果为:
p->getI() = 100
p->getI() = 40
pInt1 == p : 1
pInt2 == p : 1
- 一些建议
- 先继承自一个父类,然后实现多个接口
- 父类中提供equal( )成员函数
- equal( ) 成员函数用于判断指针是否指向当前对象
- 与多重继承相关的强制类型转换用 dynamic_cast 完成
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