Meta Stack Exchange is where users like you discuss bugs, features, and support issues that affect the software powering all 153 Stack Exchange communities.

What is meta?
Here's how it works:
  1. Any Stack Exchange user can ask a question
  2. The community provides support, votes on ideas, and reports bugs
  3. Your voice helps shape the way Stack Exchange operates

Where can I find documentation on what anonymous users are?

Expect stuff like 'users who did not register', 'users whose username is 'user34894'', 'users who can get reputation points', yadda yadda, but I couldn't locate any documentation.

share|improve this question – Bala Feb 12 '14 at 9:30
What do you want to know for "users whose username is 'user34894'"? – hims056 Feb 12 '14 at 9:37
@hims056 I want to show the documentation about how it works here to another human being. I can't just go and tell him, 'hi, you know, they name these people user345345 and, I think, authenticate them using cookies...'. – user173505 Feb 12 '14 at 9:51
@hims056: see here ( as I'd like to urge Wikimedia community to consider doing it the same way you do it, but you appear to lack documentation. – user173505 Feb 12 '14 at 9:52
It's not clear what you are looking for. Anonymous are different from registered users. People with the display name of userXXXXX are not anonymous (hence they have a name), but may be registered or unregistered. What are you trying to get at? – psubsee2003 Feb 12 '14 at 10:01
@Bata: see above for context; the page you linked is not to the specific point (although that wasn't as obvious from the original question wording). – user173505 Feb 12 '14 at 10:01
@Svetlana If that is not the specific point and not obvious from your wording can you perhaps edit your question so it becomes obvious what you want? – rene Feb 12 '14 at 10:07
You see that I'm confused, I know who I am, but not who user#### is, and not what anonymous user looks like. That's caused by lack of documentation. I think. – user173505 Feb 12 '14 at 11:09

You must log in to answer this question.