In order to use cout as such : std::cout << myObject, why do I have to pass an ostream object? I thought that was an implicit parameter.
ostream &operator<<(ostream &out, const myClass &o) { out << o.fname << " " << o.lname; return out; } Thanks
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3 Answers
You aren't adding another member function to ostream, since that would require redefining the class. You can't add it to myClass, since the ostream goes first. The only thing you can do is add an overload to an independent function, which is what you're doing in the example.
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Only if it is a member function of the class that would otherwise be the first argument. Thus, it would be:
class ostream { ... ostream &operator << (const myClass &o); ... }; Since ostream was written long before your class, you see the problem of getting your class in there. Thus, we must implement the operator as a freestanding function:
(return type) operator << ( (left hand side), (right hand side) ); When operators are implemented as member-functions of classes, the left hand side is this, and the argument becomes the right hand side. (For binary operators - unary operators work similarly.)
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Because you are overloading a free function, not a member function.
answered Dec 3, 2010 at 16:54
Nikolai Fetissov
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