这重载操作符哪里错了？。。帮我看看谢谢~~~

[touchwhat]

#include <iostream>
using namespace std;
class testclass{
public:
  int a, b;
  testclass(int a1, int b1){
    a = a1, b = b1;
  }
  
  friend ostream & operator <<(ostream & os, testclass a1){
    cout << "a = " << a1.a << ", b = " << a1.b << endl;
    return os;
  }
  
  friend ostream & operator <<(ostream & os, testclass *a1){
    cout << "a = " << a1->a << ", b = " << a1->b << endl;
    return os;
  }

  friend istream& operator >>(istream &in,const testclass &b1)
  {
	  in>>b1.a;
	  return in;
  }
  friend ostream & operator <<(ostream & os, testclass *b1){
	  cout << "b1="<<b1.a<<endl;
  return os;
  }
};//end of testclass

int main(){
  testclass *a1 = new testclass(1, 2);
  cout << a1;
  cout << *a1;
 testclass b3(4,5);
  cin>>b3;
  cout<<b3;
  
  return 0;
}

[lgccaa]

报什么错？把错误信息贴出来

[downmooner]

cout cin 换成你的形参

[downmooner]

函数里的.

[mengde007]

friend istream& operator >>(istream &in,const testclass &b1)
{
in>>b1.a;//b1都为const 了，那还能操作啊！把const去掉
return in;
}
testclass *a1 = new testclass(1, 2);
//似乎没有delete;

[douyangyang]

有两个问题：

1.你定义的第二个运算符和你的第四个运算符重复了
friend ostream & operator <<(ostream & os, testclass *a1) {...}

friend ostream & operator <<(ostream & os, testclass *b1) {...}

这两个只有参数的名字不一样，这是重复定义

2.第三个运算符
friend istream& operator >>(istream &in,const testclass &b1) {...}
应该把第二个参数的const去掉，因为你会在其中修改b1的值。

[猫已经找不回了]

friend istream& operator >>(istream &in,const testclass &b1)
{
in>>b1.a;
return in;
}
应该是这里了，const对象接收流

[liusichen_0]

这里还有一点问题
[code=c/c++]
friend ostream & operator <<(ostream & os, testclass *b1){
cout << "b1="<<b1.a<<endl; // 怎么不是b1->a呢
[/code]

[liusichen_0]

[leewon1988]

正解
[Quote=引用 5 楼 douyangyang 的回复:]
有两个问题：

1.你定义的第二个运算符和你的第四个运算符重复了
friend ostream & operator < <(ostream & os, testclass *a1) {...}

friend ostream & operator < <(ostream & os, testclass *b1) {...}

这两个只有参数的名字不一样，这是重复定义

2.第三个运算符
friend istream& operator >>(istream &in,const testclass &b1) {...}
应该把第二个参数的const去掉，因为你会在其中修改b1的值。
[/Quote]