Understanding operator << overloading

[MichaelKnight]

#include <iostream>
using namespace std;

int & fonksiyon(int&);
class Date
{
int mo, da, yr;
public:
Date(int m, int d, int y)
{
mo = m; da = d; yr = y;
}
friend ostream& operator<<(ostream& os, const Date& dt);
};

ostream& operator<<(ostream& os, const Date& dt)
{
os << dt.mo << '/' << dt.da << '/' << dt.yr;
return os;
}

int main()
{
Date dt(5, 6, 92);
cout <<"Date:" << dt << ".";
}


I thought about this thing much,and searched for a good explanation on web too,but I couldn't find what I need,and I'm bored.

The thing which I don't understand is:
ostream& operator<<(ostream& os, const Date& dt);

ostream& os is a reference for cout I think ,operator << function takes cout as an argument I think.

Then what does :
return os;

do?

[weaknessforcats]

Here you go:

cout << a << b << c;
 

Everything in C++is a function. Above are three calls to operator<< in one statement. The above can be rewritten as:

operator<<(operator<<(operator<<(cout, a), b), c);
 
 

This is not very readable but you can see a call to operator<<(cout,a) must return cout for the call to operator<<cout,b). If your operator<< does not return an ostream& then you can't use it in nested examples like this.

[MichaelKnight]

Thanks for your clear answer :)