C++重载一些需要注意的地方
1：
我一般习惯于把<<操作符重载成friend函数
class station
{
public:
string m_info;
station(string info):m_info(info){cout<<"construct!"<<endl;};
~station(){cout<<"destruct!"<<endl;};
friend ostream &operator <<(ostream &os,station & sta);
};
ostream &operator <<(ostream &os,station & sta)
{
cout<<sta.m_info;
return os;
}
这样子写是可以的
2：
将自定义类型存放于set等STL容器中时 需要重载<操作符
这里需要注意格式：
class station
{
public:
string m_info;
station(string info):m_info(info){cout<<"construct!"<<endl;};
~station(){cout<<"destruct!"<<endl;};
friend ostream &operator <<(ostream &os,station & sta);
bool operator <(const station &temp)const //这里需要注意了 ！
{
return this->m_info.compare(temp.m_info);
}
};
我刚开始写的是
bool operator <(station &temp)
{
return this->m_info.compare(temp.m_info);
}
结果无法编译通过
报错： error: no match for 'operator<' in '__x < __y'
这里 我想 set等ＳＴＬ容器在使用＜的重载函数声明 时 声明的就是 bool operator < (const TYPE &) const;
集成的时候显然也要符合要求
如果只是自己使用：
class station
{
public:
string m_info;
station(string info):m_info(info){cout<<"construct!"<<endl;};
~station(){cout<<"destruct!"<<endl;};
friend ostream &operator <<(ostream &os,station & sta);
bool operator <(station &temp)
{
return this->m_info.compare(temp.m_info);
}
};
station a("wangshuai");
station b("minhua");
cout<<(a<b)<<endl;
则显然不需要这种函数声明
3:
set <station*>的话 其实不需要重载<运算符
因为它是按照指针的大小（业绩指针在内存中的位置）比较的。。。
set<station *> stationList;
stationList.insert(new station("cantaloupes"));
stationList.insert(new station("orange"));
stationList.insert(new station("apple"));
stationList.insert(new station("banana"));
stationList.insert(new station("grapes"));
stationList.insert(new station("grapes"));
for(set<station* >::iterator itor= stationList.begin();itor!=stationList.end();itor++)
{
cout<<*(*itor)<<" "<<*itor<<endl;
}
orange 0x330f48
apple 0x330f98
banana 0x330fe8
grapes 0x331038
grapes 0x331088
cantaloupes 0x332f80
4：
看这篇文章吧
