Heavy operator <

When I learned "Deep Exploration", I found that the Code provided in the original article is roughly as follows (3rd pages in the book)

1 class Point3d2 {3     inline ostream&4     operator <<(ostream& os, const Porint3d &pt)5     {6         ...7     } 8 };

It cannot be compiled (G ++ ). Actually, the compile _ compiler is a bit difficult. I thought it was possible to compile it. Actually, practice is the only criterion for verifying truth. A bunch of queries on the Internet found that there are roughly two types of statements:

1. The input and output operators are generally overloaded as friend functions ,...
2. Binary operators can be reloaded as friend functions or member functions ,...

I can't help but wonder why. I didn't have to say when or when. Okay, come on.

 1 #include <iostream> 2 using namespace std; 3  4 class Point3d 5 { 6 private: 7     double x; 8     double y; 9     double z;10 public:11     Point3d(double tx, double ty, double tz);12     13     friend ostream&14     operator <<(ostream& os, const Point3d &pt);15 16     inline ostream&17     operator <<(ostream& os)18     {19         os<<"inline method, "<<"x: "<<x<<", y: "<<y<<", z: "<<z;20         return os;21     }22 };23 24 ostream&25 operator <<(ostream& os, const Point3d &pt)26 {27     os<<"friend method, "<<"x: "<<pt.x<<", y: "<<pt.y<<", z: "<<pt.z;28     return os;29 }30 31 Point3d::Point3d(double tx, double ty, double tz)32 {33     x=tx;34     y=ty;35     z=tz;36 }37 38 int main(void)39 {40     Point3d a(1,2,3);41     cout<<a<<endl;42     a<<cout<<endl;43 }

Looking at the proud look of the 42nd lines of code, I can only kneel down. The output result is as follows:

1 friend method, x: 1, y: 2, z: 32 inline method, x: 1, y: 2, z: 3

In a single sentence, the input operator >>and the output operator <, can be a member of a friend.

Heavy operator <