自定义输入输出流的那些事儿

今天，帮同学调了调代码，发现网上关于自定义输出输入流的问题太少了，以下就我同学的代码简单的谈一谈吧~

以下是我同学的错误代码：

#include<iostream>
using namespace std;
class point{
int x,y;
public:
void set(int a,int b){x=a;y=b;}
point operator+(const point&d)
{
point s;
s.set(x+d.x,y+d.y);
return s;
}
friend ostream & operator<<(ostream & o,const  point&d);
friend istream & operator>>(istream & is,const  point&d);
};
inline ostream&operator<<(ostream & o,const point&d)
{
return o<<"("<<d.x<<","<<d.y<<")"<<endl;;
}
inline istream&operator>>(istream & is,const point&d)
{
return is>>d.x>>d,y;
}
int main()
{
point s,t;
cout<<"please enter two numbers"<<endl;
cin>>s;
cout<<"please enter another two numbers"<<endl;
cin>>t;
cout<<"the total is"<<"   ";
cout<<s+t;
return 0;
}

我用的IDE是code::blocks,报错信息为：

error:ambiguous overload for 'operater>>' in 'is>>d.point::x'

以下是这份代码有问题的几个地方：

1.友元(friend)不需要定义inline类型；

2.point前不需要const定义；

3.友元定义的输入输出流模板有误：

应该为：friend ostream & operator<<(ostream &;point &);

friend istream & operator>>(istream &;point &);

4.问题最大的还是cout<<s+t;

这段代码直接违背了自己写的自定义输出流。

应该改为:

point c;

c=s+t;

cout<<c;

以下是改正后的代码：

#include<iostream>
using namespace std;
class point{
int x,y;
public:
void set(int a,int b){x=a;y=b;}
point operator+(point&d)
{
point s;
s.set(x+d.x,y+d.y);
return s;
}
friend ostream & operator<<(ostream & ,  point&);
friend istream & operator>>(istream & ,  point&);
};
ostream&operator<<(ostream & o, point&d)
{
return o<<"("<<d.x<<","<<d.y<<")"<<endl;

}
istream&operator>>(istream & is, point&d)
{

return is>>d.x>>d.y;
}
int main()
{
point s,t,c;
cout<<"please enter two numbers"<<endl;
cin>>s;
cout<<"please enter another two numbers"<<endl;
cin>>t;
cout<<"the total is"<<"   ";
c=s+t;
cout<<c;
return 0;
}

以下是跑出来的结果：



希望能够帮助大家！