Why does operator>> (or <<) overloading func

Question

From cplusplus.com, I saw that ostream class's member function operator<< looks like this:

ostream& operator<< (bool val);   ostream& operator<< (int val);   

.... and so on.

It does make sense because when you use the Cout object like cout<<x you activate the ostream& operator<< (int val) function, so you actually use the << operator on Cout object. This is very much like every other operator and sends the int variable to the function. What is the difference and what exactly happens when I want to stream an object of my own? Why does the syntax is suddenly ostream& operator<< (**ostream &os**, object ob)? Why do I need to add the ostream var? I am still using cout<<ob so whay isnt it just ostream& operator<< (object obj)? All I pass is my object. The cout object is allready there.

Answer 1

Some of the stream extractors are members of basic_istream; because they are members, the basic_istream argument is implied. Some of the stream extractors are not members of basic_istream. Because they are not members, the basic_istream argument has to be part of the declaration.

Like this (oversimplified):

class basic_istream {
public:
    basic_istream& operator>>(int& i);
}

basic_istream& operator>>(basic_istream&, std::string& str);

Both can be called in the same way:

int i;
std::cin >> i; // calls basic_istream::operator>>(int&)
std::string str;
std::cin >> str; // calls operator>>(basic_istrea&, std::string&)

Answer 2

operator<< is generally defined as a free function; that is, not a member function. Since it is an overloaded binary operator, that means it get's its left argument first and its right argument second.

The operator<< traditionally returns a reference to its left argument to enable the idiomatic chain of output.

To make it obvious to the reader, I tend to define my operator overloads using the lhs and rhs abbreviations; an operator<< would look similar to this, for some type T.

std::ostream& operator<<(std::ostream& lhs, T const& rhs)
{
    // TODO: Do something
    return lhs;
}

As a member function

As with other binary it could be defined as a member function. That is, let us suppose that you with defining your own iostream. Amongst other things, your class declaration may look like this. Again, T is a particular type.

class MyIOStream : public iostream
{

public:
    MyIOStream& operator<<(T const& rhs);

}

In this case operator<< is a member function. It has the same semantics when used as <<.

References

1. Operators in C and C++ - a great summary of all the operators you can overload and their typical arguments.

Answer 3

why do I need to add the ostream var?

I'm sure it's there so that you can chain outputs together:

cout << foo << bar

The first call, cout << foo will result in an ostream reference that can be used for the << bar part.