Understanding overloading of << operator

[Awareness]

#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?

[Alpo]

I could be wrong, but I believe the purpose of returning the ostream& is so that it can be used in intermediate evaluations when you have many data types tied together with << operators.

For example, try returning void. Chances are everything will print correctly so long as you are only printing one thing ( std::cout << X; ). When you need to print many things ( std::cout << X << "\n" << Y; ), you will need the operator to return the reference.

[Awareness]

I could be wrong, but I believe the purpose of returning the ostream& is so that it can be used in intermediate evaluations when you have many data types tied together with << operators.

For example, try returning void. Chances are everything will print correctly so long as you are only printing one thing ( std::cout << X; ). When you need to print many things ( std::cout << X << "\n" << Y; ), you will need the operator to return the reference.

Thanks for your answer.What does ostream& return actually?

 cout  <<"Date:" << dt << "."; 

Does it become like

cout << "Date:" << cout << "."

when it returns?

[jimblumberg]

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

You may want to study this link.

Jim

[Alpo]

Thanks for your answer.What does ostream& return actually?

 cout  <<"Date:" << dt << "."; 

Does it become like

cout << "Date:" << cout << "."

when it returns?

Taking an expression like

 std::cout << A << B << C;

I think you can look at it something like:

1.

( ( ( std::cout << A ) << B ) << C ); // If you look at the () like a function

2.

( ( std::cout << B ) << C ); // The std::cout you see here would be the return of #1

3.

( std::cout << C );

[Elysia]

You may want to study this link.

Jim

Your link is broken.

Anyway, the expression (a << b) is translated to operator << (a, b) or a.operator << (b) by the compiler. Then it just looks for an overload as usual.
So the function says, give me anything where a is of type std::ofstream or any of its derived types and where b is a Date or any of its derived classes.
std::ostream is the base class for all output streams. It can be a file stream, a string stream, or yes, even a stream to the console or anything else than derived from std::ostream. In your case, "a" is std::cout and "b" is dt. Make sense?

About the return: Alpo is correct.
If you write

std::cout << A << B << C;

It is interpreted by the compiler as

( ( ( std::cout << A ) << B ) << C ); // If you look at the () like a function

So we basically call operator << for the return of the express (std::cout << A). For that reason, we tend to return the stream itself. So if we call operator << with std::cout, then we return std::cout so it becomes

( ( std::cout << B ) << C ); // If you look at the () like a function

Of course, [I]this works with any type of stream,[/I] not just std::cout.

[Awareness]

Thanks for your answers.