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语法如下:
class Base{
};
class Name:access-specifier Base{
//access-specifier可以是public、private、protect
};
一个简单的继承结构程序:
#include <iostream> using namespace std; // 定义一个基类Fish class Fish { public: bool isFreshWaterFish; // 一个布尔型成员变量,表示鱼是淡水鱼还是咸水鱼 // 一个虚函数Swim,根据鱼的种类输出不同的游泳场所 void Swim() { if (isFreshWaterFish) { cout << "swim in lake" << endl; // 如果是淡水鱼,输出在湖中游泳 } else { cout << "swim in sea" << endl; // 如果是咸水鱼,输出在海中游泳 } } }; // 定义一个派生类Tuna,继承自Fish class Tuna : public Fish { public: Tuna() { isFreshWaterFish = false; // 将isFreshWaterFish设置为false,表示Tuna是咸水鱼 } }; // 定义另一个派生类Carp,同样继承自Fish class Carp : public Fish { public: Carp() { isFreshWaterFish = true; // 将isFreshWaterFish设置为true,表示Carp是淡水鱼 } }; int main() { Carp lunch; // 创建一个Carp类型的对象lunch Tuna dinner; // 创建一个Tuna类型的对象dinner cout << "lunch: " << endl; // 输出午餐的描述 lunch.Swim(); // 调用lunch的Swim函数 cout << "dinner: " << endl; // 输出晚餐的描述 dinner.Swim(); // 调用dinner的Swim函数 system("pause"); return 0; }
将属性声明为protected时,允许友元类和派生类访问该属性,但禁止在继承结构之外访问
示例代码如下
class Fish
{
protected://声明为protected
bool isFreshWaterFish;
public:
void Swim(){
if (isFreshWaterFish)
{
cout<<"swim in lake"<<endl;
}
else{
cout<<"swim in sea"<<endl;
}
}
};
如果基类包含重载的构造函数,可以使用初始化列表传递参数
示例代码如下:
#include <iostream> using namespace std; // 定义基类Fish class Fish { protected: bool isFreshWaterFish; // 一个受保护的布尔型成员变量,表示鱼是淡水鱼还是咸水鱼 Fish(bool isFreshWater) : isFreshWaterFish(isFreshWater) {} // 构造函数,根据传入的参数初始化isFreshWaterFish public: // 一个虚函数Swim,根据鱼的种类输出不同的游泳场所 void Swim() { if (isFreshWaterFish) { cout << "swim in lake" << endl; // 如果是淡水鱼,输出在湖中游泳 } else { cout << "swim in sea" << endl; // 如果是咸水鱼,输出在海中游泳 } } }; // 定义派生类Tuna,继承自Fish class Tuna : public Fish { public: // Tuna的构造函数,使用Fish的构造函数并传入false,表示Tuna是咸水鱼 Tuna() : Fish(false) {} }; // 定义另一个派生类Carp,同样继承自Fish class Carp : public Fish { public: // Carp的构造函数,使用Fish的构造函数并传入true,表示Carp是淡水鱼 Carp() : Fish(true) {} }; int main() { Carp lunch; // 创建一个Carp类型的对象lunch Tuna dinner; // 创建一个Tuna类型的对象dinner cout << "lunch: " << endl; // 输出午餐的描述 lunch.Swim(); // 调用lunch的Swim函数 cout << "dinner: " << endl; // 输出晚餐的描述 dinner.Swim(); // 调用dinner的Swim函数 system("pause"); return 0; }
这样可以强迫在构造一个派生类时,必须定义一个这样的函数来为其赋值,避免不定义的情况下,随机数值产生影响
只要派生类中实现了从基类中继承的函数,且返回值和标值相同,就可以覆盖
class Fish { protected: bool isFreshWaterFish; Fish(bool isFreshWater):isFreshWaterFish(isFreshWater){}; public: void Swim(){ if (isFreshWaterFish) { cout<<"swim in lake"<<endl; } else{ cout<<"swim in sea"<<endl; } } }; class Tuna:public Fish{ public: Tuna():Fish(false){}; void Swim(){//覆盖基类中的方法 cout<<"fu gai"<<endl; } };
使用作用域解析运算符::
Dinner.Fish::Swim();//调用基类中的方法
Dinner.Swim();//调用覆盖后的方法
同样使用作用域解析运算符::
class Fish { protected: bool isFreshWaterFish; Fish(bool isFreshWater):isFreshWaterFish(isFreshWater){}; public: void Swim(){ if (isFreshWaterFish) { cout<<"swim in lake"<<endl; } else{ cout<<"swim in sea"<<endl; } } }; class Tuna:public Fish{ public: Tuna():Fish(false){}; void Swim(){//覆盖基类中的方法 cout<<"fu gai"<<endl; Fsih::Swim();//调用基类方法 } };
覆盖是一种极端情况,即派生类中的重载方法会隐藏基类中方法的所有重载版本,导致需要调用时无法调用,示例代码如下
#include <iostream> using namespace std; // 定义基类Fish class Fish { public: // 定义一个成员函数Swim void Swim() { cout << "Fish swim" << endl; } // 重载Swim函数,带有一个布尔参数isFreshWaterFish void Swim(bool isFreshWaterFish) { if (isFreshWaterFish) { cout << "swim in lake" << endl; } else { cout << "swim in sea" << endl; } } }; // 定义派生类Tuna, class Tuna : public Fish { public: // 重写基类Fish的Swim成员函数 void Swim() { cout << "fu gai" << endl; } }; int main() { Tuna dinner; // 创建Tuna类的对象dinner cout << "dinner: " << endl; dinner.Swim(); // 调用dinner对象的Swim函数,由于Tuna类重写了Swim,所以调用的是Tuna类的Swim dinner.Swim(false); //ERROR,因为派生类中的重载函数Swim(bool isFreshWaterFish)已经被覆盖了 system("pause"); return 0;
若想调用基类中的方法
解决办法1: 使用作用域解析符::
dinner.Fish::Swim(false);
解决方法2: 在派生类中使用using
解除隐藏
#include <iostream> using namespace std; class Fish { public: void Swim(){ cout<<"Fish swim"<<endl; } void Swim(bool isFreshWaterFish){ if (isFreshWaterFish) { cout<<"swim in lake"<<endl; } else{ cout<<"swim in sea"<<endl; } } }; class Tuna:public Fish{ public: using Fish::Swim;//解除隐藏 void Swim(){ cout<<"fu gai"<<endl; } }; int main(){ Tuna dinner; cout<<"dinner: "<<endl; dinner.Swim(); dinner.Swim(false);//可以正常调用了 system("pause"); return 0; }
解决方法3: 在派生类中,把基类中的方法全部重载一次,且在重载方法中就调用基类方法
#include <iostream> using namespace std; class Fish { public: void Swim(){ cout<<"Fish swim"<<endl; } void Swim(bool isFreshWaterFish){ if (isFreshWaterFish) { cout<<"swim in lake"<<endl; } else{ cout<<"swim in sea"<<endl; } } }; class Tuna:public Fish{ public: //重载基类中的方法Swim() void Swim(){ cout<<"fu gai"<<endl; } //重载基类中的重载方法Swim(bool isFreshWaterFish) void Swim(bool isFreshWaterFish){ Fish::Swim(isFreshWaterFish);//直接调用基类中的方法 } }; int main(){ Tuna dinner; cout<<"dinner: "<<endl; dinner.Swim(); dinner.Swim(false);//可以正常运行了 system("pause"); return 0; }
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