Overloading operator "<<"

[Sunny]

The way to overload operator << is : ostream& operator << (ostream&
os, const Obj& obj);
and this is a member function.
My question is why do we need to provide a const reference of Obj as
argument when it is a member function and its members accessible
through this pointer ?
Isnt cout << obj;
equivalent to obj.operator<<( cout) ?
Thanks

[Robert Bauck Hamar]

Re: Overloading operator &quot;&lt;&lt;& quot;

Sunny wrote:

The way to overload operator << is : ostream& operator << (ostream&
os, const Obj& obj);

Yes

and this is a member function.

No

My question is why do we need to provide a const reference of Obj as
argument when it is a member function and its members accessible
through this pointer ?

It's not a member of Obj.

Isnt cout << obj;
equivalent to obj.operator<<( cout) ?

No, it is equivalent to either of

cout.operator<< (obj);
operator<<(cout , obj);

--
Robert Bauck Hamar
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