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一、 C++成员函数的重载
C++中的成员函数有四种,分别是普通成员函数,virtual虚函数,const成员函数。
  • (1) void func(int a);
  • (2) virtual void func(int a);
  • (3) void func(int a) const;
如果在一个类中,声明这四种函数,哪是重复定义?哪些是重载?
其中(1)(2)是重复定义,故编译不能通过,而(3)与(1)(2)是不同类型的函数,是重载。
成员函数被重载的特征是:
  • (1)具有相同的作用域(即同一个类定义中);
  • (2)函数名字相同
  • (3)参数类型,顺序 或 数目不同(包括const参数和非const函数)
  • (4)virtual关键字可有可无。
从成员函数的重载特征中,可以知道(1)(2)是重复定义。那么(3)为什么和(1)(2)不同呢?
因为类中的函数,都会自动添加一个自身类指针this,所以
void func(int a) ==== void func(Base * this, int a)virtual func(int a) ==== virtual func(Base *this, int a)void func(int a)const === void func(const Base *this, int a) const
所以(3)可以与(1)(2)发生重载,因为参数有一个const。
二 、C++成员函数的继承#include <iostream> using namespace std; class Base { public:     void f(int a){         cout << "Base::f(int a)" << endl;     }     virtual void g(int a) {         cout << "virtual Base::g(int a)" << endl;     } }; class Derived : public Base { public:     void h(int a) {         cout << "Derivd::h(int a)" << endl;     } }; int main() {     Base b;     b.f(3);     b.g(4);     Derived d;     d.f(3);     d.g(4);     d.h(3); } #include <iostream>using namespace std;class Base {public: void f(int a){  cout << "Base::f(int a)" << endl; } virtual void g(int a) {  cout << "virtual Base::g(int a)" << endl; }};class Derived : public Base{public: void h(int a) {  cout << "Derivd::h(int a)" << endl; }};int main(){ Base b; b.f(3); b.g(4); Derived d; d.f(3); d.g(4); d.h(3);}
Base b的对象模型:
Derived d的对象模型:
则在子类Derived d中继承了父类中的virtual void g(int a) ; void f(int a);
运行结果为:

三、C++成员函数的覆盖
覆盖是指派生类重新实现(或者改写)了基类的成员函数,其特征是:
  • (1)不同的作用域(非别位于派生类和基类中);
  • (2)函数名称相同
  • (3)参数列表完全相同;
  • (4)基类函数必须是虚函数。
从(4)中我们得知覆盖只是针对虚函数的。
#include <iostream> using namespace std; class Base { public:     void f(int a){         cout << "Base::f(int a)" << endl;     }     virtual void g(int a) {         cout << "virtual Base::g(int a)" << endl;     } }; class Derived : public Base { public:     void h(int a) {         cout << "Derivd::h(int a)" << endl;     }     virtual void g(int a) {         cout << "virtual Derived::g(int a)" << endl;     } }; int main() {     Base b;     b.f(3);     b.g(4);     Derived d;     d.f(3);     d.g(4);     d.h(3); } #include <iostream>using namespace std;class Base {public: void f(int a){  cout << "Base::f(int a)" << endl; } virtual void g(int a) {  cout << "virtual Base::g(int a)" << endl; }};class Derived : public Base{public: void h(int a) {  cout << "Derivd::h(int a)" << endl; } virtual void g(int a) {  cout << "virtual Derived::g(int a)" << endl; }};int main(){ Base b; b.f(3); b.g(4); Derived d; d.f(3); d.g(4); d.h(3);}
Derived d对象模型如下:
其中Derived中重新定义了基类的虚成员函数virtual void g(int a);
四、C++成员函数的隐藏
隐藏是指派生类的成员函数遮蔽了与其同名的基类成员函数,具体规则如下:
(1) 派生类的函数与基类的函数同名,但是参数列表有所差异。此时,不论有无virtual关键字,基类的函数在派生类中将被隐藏。(注意别与重载混合)
(2)派生类的函数与基类的函数同名,参数列表也相同,但是基类函数没有virtual关键字。此时,基类的函数在派生类中将被吟唱。(注意别与覆盖混合)
判断下面哪些函数是覆盖,哪些函数是隐藏?
#include <iostream> using namespace std; class Base { public:     virtual void f(float x) {         cout << "virtual Base::f(float) " << x << endl;     }     void g(float x) {         cout << "Base::g(float) " << x << endl;     }     void h(float x) {         cout << "Base::h(float) " << x << endl;     } }; class Derived : public Base{ public:     virtual void f(float x) {         cout << "virtual Derived::f(float) " << x << endl;     }     void g(int x) {         cout << "Derived::g(int) " << x << endl;     }     void h(float x) {         cout << "Derived::h(float) " << x << endl;     } }; int main(void) {     Derived d;     Base *pb = &d;     Derived *pd = &d;     pb->f(3.14f);     pd->f(3.14f);     pb->g(3.14f);     pd->g(3.14f);     pb->h(3.14f);     pd->h(3.14f); } #include <iostream>using namespace std;class Base {public: virtual void f(float x) {  cout << "virtual Base::f(float) " << x << endl; } void g(float x) {  cout << "Base::g(float) " << x << endl; } void h(float x) {  cout << "Base::h(float) " << x << endl; }};class Derived : public Base{public: virtual void f(float x) {  cout << "virtual Derived::f(float) " << x << endl; } void g(int x) {  cout << "Derived::g(int) " << x << endl; } void h(float x) {  cout << "Derived::h(float) " << x << endl; }};int main(void){ Derived d; Base *pb = &d; Derived *pd = &d; pb->f(3.14f); pd->f(3.14f); pb->g(3.14f); pd->g(3.14f); pb->h(3.14f); pd->h(3.14f);}
其中子类Derived中 vitual void f(float x)  是覆盖,而void g(int x) 和void h(float x)都是隐藏。
运行结果:
再看一个例子:
#include <iostream> using namespace std; class Base { public:     virtual void f(int a) {         cout << "virtual Base::f(int a)" << endl;     }     void f(double d) {         cout << "Base::f(double d)" << endl;      } }; class Derived : public Base { public:     void f(double d) {         cout << "Derivd::f(double d)" << endl;     } }; int main() {     Derived d;     d.f(3);     d.f(2.5);     Derived *pd = new Derived();     pd->f(3);     pd->f(2.5);     Base b;     b.f(5);     b.f(3.5);     Base *pBase = new Derived();     pBase->f(5);     pBase->f(3.5); } #include <iostream>using namespace std;class Base{public: virtual void f(int a) {  cout << "virtual Base::f(int a)" << endl; } void f(double d) {  cout << "Base::f(double d)" << endl; }};class Derived : public Base{public: void f(double d) {  cout << "Derivd::f(double d)" << endl; }};int main(){ Derived d; d.f(3); d.f(2.5); Derived *pd = new Derived(); pd->f(3); pd->f(2.5); Base b; b.f(5); b.f(3.5); Base *pBase = new Derived(); pBase->f(5); pBase->f(3.5);}
其中父类中的void f(double d)隐藏了子类的virtual void f(int a),  void f(double d)函数。
所以在主函数中
Derived d; d.f(3); d.f(2.5); Derived *pd = new Derived(); pd->f(3); pd->f(2.5); Derived d;d.f(3);d.f(2.5);Derived *pd = new Derived();pd->f(3);pd->f(2.5);只要通过Derived对象或者Derived指针执行f()函数,都只执行void Derived::f(double d)该函数。[html]Base *pBase = new Derived(); pBase->f(5); pBase->f(3.5); Base *pBase = new Derived();pBase->f(5);pBase->f(3.5);
在调用pBase->f(5)时,首先要去pBase类中找到对应需要执行的函数,因为Base类中有两个函数virtual void f(int a) 和 void f(double)重载,因为该实参是5,为int类型,所以要调用virtual void f(int a)函数,因为该f(int a)是一个虚函数,所以再去判断pBase所指向的具体对象,具体对象为Derived子类,再去Derived子类的虚函数表中找到void f(int a)函数。因为Derived子类继承了父类Base的虚函数vitural void f(int a),所以输出 virtual Base::f(int a);
在调用pBase->f(3.5)时,首先要去pBase类中找到对应需要执行的函数,因为因为Base类中有两个函数virtual void f(int a) 和 void f(double)重载,因为该实参是3.5,为double类,所以要调用void f(double d)函数,因为该函数是一个普通成员函数,故直接输出。 void Base::f(double d);
再举一个例子:
#include <iostream> using namespace std; class Base { public:     virtual void f(int a) {         cout << "virtual Base::f(int a)" << endl;     }     void f(double d) {         cout << "Base::f(double d)" << endl;      } }; class Derived : public Base { public:     void f(int a) {         cout << "virtual Derived::f(int a)" << endl;     } }; int main() {     Derived d;     d.f(3);     d.f(2.5);     Derived *pd = new Derived();     pd->f(3);     pd->f(2.5);     Base b;     b.f(5);     b.f(3.5);     Base *pBase = new Derived();     pBase->f(5);     pBase->f(3.5); } #include <iostream>using namespace std;class Base{public: virtual void f(int a) {  cout << "virtual Base::f(int a)" << endl; } void f(double d) {  cout << "Base::f(double d)" << endl; }};class Derived : public Base{public: void f(int a) {  cout << "virtual Derived::f(int a)" << endl; }};int main(){ Derived d; d.f(3); d.f(2.5); Derived *pd = new Derived(); pd->f(3); pd->f(2.5); Base b; b.f(5); b.f(3.5); Base *pBase = new Derived(); pBase->f(5); pBase->f(3.5);}
子类Derived中的void f(int a)既覆盖了基类Base的虚函数virtual void f(int a),也隐藏了基类的virtual void f(int a),  void f(double d)函数。
Derived d; d.f(3); d.f(2.5); Derived *pd = new Derived(); pd->f(3); pd->f(2.5); Derived d;d.f(3);d.f(2.5);Derived *pd = new Derived();pd->f(3);pd->f(2.5);
同理所有用子类对象或者子类指针来调用f()函数时,都只执行virtual void f(int a),输出virtual Derived::f(int a)
view plaincopyprint?Base *pBase = new Derived(); pBase->f(5); pBase->f(3.5); Base *pBase = new Derived();pBase->f(5);pBase->f(3.5);
pBase->f(5),首先去Base类中寻找相应的函数,同理Base类中的两个函数virtual void f(int a)和void f(double d)是重载函数,因为实参为5,为int类型,所以我们要调用virtual void f(int a)虚函数,因为该函数是虚函数,所以要去判断pBase指向的具体对象,因为pBase指向的是一个子类的对象,所以需要去子类的虚函数表中取找virtual void f(int a)函数,找到之后,执行该函数,故输出virtual Derived::f(int a)。
pBase->f(3.5),首先去Base类中寻找相应的函数,同理Base类中的两个函数virtual void f(int a)和void f(double d)是重载函数,因为实参为3.5,为double类型,所以我们要调用void f(double d),因为该函数为普通成员函数,故执行输出: void Base::f(double d);

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