5-3 多级派生类的构造函数
Time Limit: 1000MS Memory limit: 65536K
题目描述
要求定义一个基类3个name(char *类型sex(char类型age(int类型创建Employee,增加两个数据成员 基本工资 int类型) 请假天数int型);为它定义初始化成员信息的构造函数,和显示数据成员信息的成员函数创建Manager;增加一个成员 业绩 );为它定义初始化成员信息的构造函数,和显示数据成员信息的成员函数共如示例数据所示,共<font face="\"Times" new="" roman,="" serif\"="" style="padding: 0px; margin: 0px;">5行,分别代表姓名、年龄、性别、基本工资、请假天数、业绩
示例输入
Jerry m 32 4200 1 100
示例输出
name:Jerry age:32 sex:m basicSalary:4200 leavedays:1 performance:100
提示
来源
黄晶晶
示例程序
#include<iostream>
#include<cstring>
using namespace std;
class person
{
protected:char name[100];char sex;
int age;
public:
person(){}
person(char *nam,char s,int a)
{
strcpy(name,nam);
sex=s;
age=a;
}
void get1()
{
cin>>name>>sex>>age;
}
void show1()
{
cout<<"name:"<<name<<endl;
cout<<"age:"<<age<<endl;
cout<<"sex:"<<sex<<endl;
}
};
class emplayee:public person
{
protected:
int basicsalary,leavedays;
public:
emplayee(){}
emplayee(char *nam,char s,int a,int b,int l):person(nam,s,a)
{
basicsalary=b;leavedays=l;
}
void get2()
{
get1();
cin>>basicsalary>>leavedays;
}
void show2()
{
show1();
cout<<"basicSalary:"<<basicsalary<<endl;
cout<<"leavedays:"<<leavedays<<endl;
}
};
class manager:public emplayee
{
protected:float performance;
public:manager(){}
manager(char *nam,char s,int a,int b,int l,float p):emplayee(nam,s,a,b,l)
{
performance=p;
}
void get3()
{
get2();
cin>>performance;
}
void show3()
{
show2();
cout<<"performance:"<<performance<<endl;
}
};
int main()
{
manager e;
e.get3();
e.show3();
return 0;
}