Ostream template overload

After a bit of googling, it seems that this is a well documented bug in the 2. Yes, everything is in one giantic. The definition of the "Key" and "Data" structures are all done within the class like those bool operators , along with the ostream operator. It is complaining about the declaration of a friend function. Tychom : My apologies for being rude.

No go. Successful instantiation results in a function, which it adds to the overload set. But how is it supposed to instantiate your function template.

It has no way of knowing what to use for T. So it doesn't instantiate template argument deduction fails , and no instance of it finds its way into the overload set. I don't see anything immediate here. Stack Overflow for Teams — Collaborate and share knowledge with a private group. Create a free Team What is Teams? Collectives on Stack Overflow. Learn more. Template overloading ostream operator Ask Question. Asked 8 years, 9 months ago. Active 8 years, 9 months ago. Viewed 4k times. Is this code illegal?

Unused template variable breaks compilation. Should it matter? EDIT: This is solved. I had to return o; 3. Doesn't compile any of them. Can someone please clarify this things for me?

Thank you. Improve this question. Define your operator in std namespace. You'll see the difference — PiotrNycz. If the compiler can't infer the type of a template argument, you have to provide it yourself.

Obviously, you can't do that when you overload an operator. When you do use T , the compiler is able to infer its type, and this is why it works. Also, your last example runs fine on ideone. Enter info1: Betty 45 Enter info2: Bill 44 Betty 45??? In addition to the missing friend mentioned by others, you have the following errors: L You're reading from cin instead of lhs. L You're writing to cout instead of lhs. And adding to that, the purpose of the stream operators is to generalize the streams whether it's to cin, cout, file input and file output streams Last edited on Jan 5, at am UTC.

Jan 5, at am UTC. JLBorges Since they take the user-defined type as the right argument b in a b , they must be implemented as non-members.

Why, please? Will you illustrate it a bit more in a simple language. Good remark. We can't go and write member functions in std::ostream , so the only option is to implement the overloaded operator as a non-member function.

Jan 5, at pm UTC. We have, e.