C++实践参考：警察与厨师

【项目-警察和厨师】
（1）根据下面的类图，定义各个类：

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

[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;
}

评论：
这样做，是不是在逻辑上很清楚了？
Person、Police类中该定义复制构造函数，在这里没有写，用其默认复制构造函数了。相关类中没有定义指针型成员，不必要深复制，所以，可以使用默认复制构造函数。