为什么“~ChildLabourer”改为“~childLabourer”才可通过
为什么要把“~ChildLabourer”改为“~childLabourer”才可通过？？？？？？？？？？？？不应该和类同名吗？不应该是大写吗？？？？？？？？？？？？？？？？？？？？？？？？？？？？？？
2 回答
我的没改就通过了，就是和类名同名，提示的应该是错误的，我的是其他方面的错误也是提示析构函数错误应小写，
如下为能运行的代码
代码如下
#include <iostream>
#include <stdlib.h>
#include <string>
using namespace std;
/**
* 定义工人类: Worker
* 数据成员: m_strName
* 成员函数: work()
*/
class Worker
{
public:
Worker(string name)
{
m_strName = name;
cout << "Worker" << endl;
}
~Worker()
{
cout << "~Worker" << endl;
}
void work()
{
cout << m_strName << endl;
cout << "work" << endl;
}
protected:
string m_strName;
};
/**
* 定义儿童类: Children
* 数据成员: m_iAge
* 成员函数: play()
*/
class Children
{
public:
Children(int age)
{
m_iAge = age;
cout << "Children" << endl;
}
~Children()
{
cout << "~Children" << endl;
}
void play()
{
cout << m_iAge << endl;
cout << "play" << endl;
}
protected:
int m_iAge;
};
/**
* 定义童工类: ChildLabourer
* 公有继承工人类和儿童类
*/
class ChildLabourer : public Worker,public Children
{
public:
ChildLabourer(string name, int age):Worker(name),Children(age)
{
cout << "ChildLabourer" << endl;
}
~ChildLabourer()
{
cout << "~ChildLabourer" << endl;
}
};
int main(void)
{
// 使用new关键字创建童工类对象
ChildLabourer *p=new ChildLabourer("workerss",19);
// 通过童工对象调用父类的work()和play()方法
p->work();
p->play();
// 释放
delete p;
p=NULL;
return 0;
}
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