C++实践参考：警察与厨师

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【项目-警察和厨师】
（1）根据下面的类图，定义各个类：
这里写图片描述
要求：
各个成员函数，只要输出相关的信息即可，暂不深究其业务功能
请为各个类增加构造函数
在实现中，可以增加需要的其他函数
自行编制main函数，完成初步的测试

[参考解答1]

#include <iostream>
using namespace std;
class Person
{public:Person(int, string);void action();string getName(){return name;}
private:int age;string name;
};
Person::Person(int a, string n):age(a), name(n) {}
void Person::action()
{cout<<name<<" do some action"<<endl;
}
class Police: public Person
{
public:Police(int, string, int);void arrest(Person);
private:int level; //级别
};
Police::Police(int a, string n, int l):Person(a,n),level(l) {}
void Police::arrest(Person p)
{cout<<" Police "<<getName()<<" arrest " <<p.getName()<<endl;
}
class Cook: public Person
{
public:Cook(int, string, double);void getCake(int);
private:double salary; //薪水
};
Cook::Cook(int a, string n, double s):Person(a,n),salary(s) {}
void Cook::getCake(int n)
{cout<<" Cook "<<getName()<<" gave me " <<n<<" cakes."<<endl;
}int main()
{Person tom(120,"Tom");Police jack(30,"Jack",2);Cook john(24,"John",5000);jack.arrest(tom);john.getCake(4);return 0;
}

（2）下面的类图，为Polic类和Cook类增加了对象成员，请扩充代码，完成上述各项要求
这里写图片描述
[参考解答1]

#include <iostream>
using namespace std;
class Person
{public:Person(int, string);void action();string getName(){return name;}
private:int age;string name;
};
Person::Person(int a, string n):age(a), name(n) {}
void Person::action()
{cout<<name<<" do some action"<<endl;
}
class Police: public Person
{
public:Police(int a, string n, int l, int la, string ln);void arrest(Person);void show();
private:int level; //级别Person leader;  //领导
};
Police::Police(int a, string n, int l, int la, string ln):Person(a,n),level(l),leader(la,ln) {}
void Police::arrest(Person p)
{cout<<"Police "<<getName()<<" arrest " <<p.getName()<<endl;
}
void Police::show()
{cout<<"Police "<<getName()<<", leader is " <<leader.getName()<<endl;
}class Cook: public Person
{
public:Cook(int a, string n, double s,int pa, string pn, int pl, int pla, string pln);void getCake(int);void show();
private:double salary; //薪水Police protector;  //厨师小店的片区警察
};
Cook::Cook(int a, string n, double s,int pa, string pn, int pl, int pla, string pln):Person(a,n),salary(s),protector(pa,pn,pl,pla,pln) {}
void Cook::getCake(int n)
{cout<<"Cook "<<getName()<<" gave me " <<n<<" cakes."<<endl;
}
void Cook::show()
{cout<<"Cook "<<getName()<<" is protected by Police "<<protector.getName()<<endl;
}int main()
{Person tom(120,"Tom");Police jack(30,"Jack",2,43,"Jerry");Cook john(24,"John",5000,30,"Jack",2,43,"Jerry");jack.show();john.show();return 0;
}

评价：
- 这些代码是完成是题目的要求，但是，并不好。
- 每个构造函数带上一长串的参数，难写，难看，这本身就是质量问题。
- 这种写法，也根本未体现对象的“封装”——都是一串散乱的基本类型数据在工作。
- 我们希望看到jack警察的上司就是一个人，john厨师的保卫者，就是一个警察。
- 需要做的是，利用对象作为构造函数的参数，使结构清晰。
- 当然，这时需要增加相关的复制构造函数了。

[参考解答2]

#include <iostream>
using namespace std;
class Person
{
public:Person(int, string);void action();string getName(){return name;}
private:int age;string name;
};
Person::Person(int a, string n):age(a), name(n) {}
void Person::action()
{cout<<name<<" do some action"<<endl;
}
class Police: public Person
{
public:Police(int a, string n, int l, Person);void arrest(Person);void show();
private:int level; //级别Person leader;  //领导
};
Police::Police(int a, string n, int l, Person p):Person(a,n),level(l),leader(p) {}
void Police::arrest(Person p)
{cout<<"Police "<<getName()<<" arrest " <<p.getName()<<endl;
}
void Police::show()
{cout<<"Police "<<getName()<<", leader is " <<leader.getName()<<endl;
}class Cook: public Person
{
public:Cook(int a, string n, double s,Police p);void getCake(int);void show();
private:double salary; //薪水Police protector;  //厨师小店的片区警察
};
Cook::Cook(int a, string n, double s,Police p):Person(a,n),salary(s),protector(p) {}
void Cook::getCake(int n)
{cout<<"Cook "<<getName()<<" gave me " <<n<<" cakes."<<endl;
}
void Cook::show()
{cout<<"Cook "<<getName()<<" is protected by Police "<<protector.getName()<<endl;
}int main()
{Person jerry(43,"Jerry");Police jack(30,"Jack",2,jerry);Cook john(24,"John",5000,jack);jack.show();john.show();return 0;
}