类的继承中带参数的构造函数

测试代码：

#include<iostream>
using namespace std;

class base1{
protected:
 int m;
public:
 base1(int x){m=x;cout<<"base1 created"<<endl;}
 ~base1(){cout<<"base1 deleted/n";}
};

class base2{
protected:
 int n;
public:
 base2(int x){n=x;cout<<"base2 created"<<endl;}
 ~base2(){cout<<"base2 deleted/n";}
};

class derive:public base1,public base2{
protected:
 int l;
public:
 derive(int x,int y,int z) :base1(x),base2(y){
  l=z;
 }
 void show(){cout<<"m is "<<m<<endl<<"n is "<<n<<endl<<"l is "<<l<<endl;}
};

int main()
{
 derive cl(1,2,3);
 cl.show();
 return 0;
}

 

测试结果：

base1 created
base2 created
m is 1
n is 2
l is 3
base2 deleted
base1 deleted
请按任意键继续. . .