为什么C ++运算符重载必须包含“ ostream＆os”？

Question

// The original code is from link: http://hoeven.blogbus.com/logs/37324287.html
#include<iostream>
#include<vector>
#include <stdlib.h>

using namespace std;

class test{
public:
     int v;
   /*构造函数*/
     test():v(0){}
     test(const int &a):v(a){}
     test(const test &t1):v(t1.v){}

   /*以下重载小于号 < */
     //比较两个对象的大小
     bool operator<(const test &t1) const{
         return (v < t1.v);
     }
     //比较对象和int的大小
     bool operator<(const int &t1) const{
         return (v < t1);
     }
     //友元函数，比较int和对象的大小
     friend inline bool operator<(const int &a, const test & t1){
         return (a < t1.v);
     }

   /*以下重载赋值号 = */
     //对象间赋值
     test & operator=(const test &t1){
         v = t1.v;
         return *this;
     }
     //int赋值给对象
     test & operator=(const int &t1){
         v = t1;
         return *this;
     }

   /*以下重载加号 + */
     //对象加上 int
     test operator+(const int & a){
         test t1;
         t1.v = v + a;
         return t1;
     }
     //对象加对象
     test operator+(test &t1){
         test t2;
         t2.v = v + t1.v;
         return t2;
     }

   /*以下重载加等号 += */  
     //对象加上对象
     test &operator+=(const test &t1){
         v += t1.v;
         return *this;
     }  
     //对象加上int
     test &operator+=(const int &a){
         v += a;
         return *this;
     }

   /*以下重载双等号 == */  
     //对象==对象
     bool operator==(const test &t1)const{
         return (v == t1.v);
     }  
     //对象==int
     bool operator==(const int &t1)const{
         return (v == t1);
     }  

   /*以下重载 输入>> 输出<< */
     /*友元函数，输出对象*/
     friend inline ostream & operator << (ostream & os, test &t1){
         cout << "class t(" << t1.v << ")" << endl;
         return os;
     }
     /*友元函数，输入对象*/
     friend inline istream & operator >> (istream & is, test &t1){
         cin >> t1.v;
         return is;
     }
};

int main(){
     test t0, t1(3);  // t0 has no initial value, so use default value 0
     test t2(t1);
     cout << t0 << t1 << t2;
     cin >> t1;
     t2 = t1;
     t2 += t1;
     t1 += 10;
     cout << t2;
     if(t1 < t2) cout << "t1 < t2";
     else if(t1 == t2) cout << "t1 = t2";
     else /* t1 > t2*/ cout << "t1 > t2";
     cout <<endl;
     system("echo Tom");
     return 0;
}

/*
 $ ./a.out 
 class t(0)
 class t(3)
 class t(3)
 45
 class t(90)
 t1 < t2
 Tom
 */

The complete code is above. But I don't understand why "ostream & os" (see below) must be there in the bracket? If I remove "ostream & os", a lot of errors were given.

 friend inline ostream & operator << (ostream & os, test &t1){
     cout << "class t(" << t1.v << ")" << endl;
     return os;
 }

Answer 1

Because the code you paste is wrong.

The << operator should write to its param. ie your function should be

inline ostream & operator << (ostream &os, test &t1) {
    os << "class t(" << t1.v << ")" << endl;
    return os;
}

This gives you the ability to write to any ostream, and not only cout . In the previous code, if you wrote ofile << t1; (considering ofile is a filestream) this would not have write to the file, but still on standard output.

Answer 2

You have 2 ways of overloading operator<< :

1. unary ( this->operator<<(other_object) ),
2. binary ( operator<<(ostream& os, other_object ).

Only one of them is possible (second), first is impossible because to implement it you have to overload function of ostream class operator<< which cant be done.
When you write statement like cout << "some text"; binary operator is invoked , which gets as arguments: object of class ostream and object of another class

Answer 3

The ostream & os is the left hand side of the operator << . ie cout << myObject; . In this case, cout is the ostream . However, in your overload, you are ignoring the os variable and using cout unconditionally. You should replace cout with os in your operator << overload.

Answer 4

Only if you pass the ostream variable to the method will the method know where to output. You could alternatively open a file stream and pass it to the method. Then the output would be written to the file instead of to standard output.

Answer 5

The ostream argument gives you further flexibility as to where to direct your output. It also allows chaining so that multiple items can be output in the same statement:

test t1;
test t2(5);
test t3(-4);

// Write "0 5 -4\n" to standard output
cout << t1 << " " << t2 << " " << t3 << endl;

// Write "0 5 -4\n" to the file my_file.txt
ofstream ofs("my_file.txt");
ofs << t1 << " " << t2 << " " << t3 << endl;

// Write "0 5 -4\n" to oss, which is then written to standard output
ostringstream oss;
oss << t1 << " " << t2 << " " << t3 << endl;
cout << oss.str() << endl;