对于这个朋友重载运算符 function 而不仅仅是 std::cout，std::ostream& out 的目的是什么？

Question

Can someone explain the purpose of std::ostream &out for this friend overloaded operator function? Why not just use std::cout inside the function?

friend std::ostream& operator<<(std::ostream& out, const Complex& c) 
{
    out << c.real;
    out << "+i" << c.imag << endl;

    return out;
}

Instead, like so:

friend operator<<(const Complex& c) 
{ 
    std::cout << c.real;
    std::cout << "+i" << c.imag << endl;
}

Answer 1

Because this way you can use any ostream as output, not just std::cout for example, std::clog << c;

Answer 2

Displaying the output in a function and returning the output stream to the overloaded friend function have different meanings.

You can pass the initialized std::ostream& to any other ostream& . For example, std::cout , std::clog , std::cerr .

Note that this can be passed to the file stream instances such as std::ofstream when writing to a file. ¹

Also, in your second given example:

friend operator<<(const Complex &c)

You are missing a return type. The program will fail to compile. Also, two arguments are required to overload the operator<< .

---

^{1. Thanks to @Pete Becker for suggesting this important information.}

Answer 3

This:

 friend std::ostream & operator << (std::ostream &out, const Complex &c) { out << c.real; out << "+i" << c.imag << endl; return out; }

is the function that gets called when you write:

Complex c;
std::cout << c;    // same as: operator<<( std::cout, c);

Calling std::cout << c; inside the implementation of operator<< would be bad, because then it would call itself recursively with no end.