what is pts/0 , :0 and attributes before it + ,? etc

While reading about Linux, I got a $ who -a, so before trying that I logged into three of my text terminals (tty1,tty2,tty3) respectively,, and then I came back to X-window (Ctrl + Alt +f7),, then I tried:-

$ who
anupam tty2 2014-09-20 16:19
anupam tty3 2014-09-20 16:20
anupam tty1 2014-09-20 16:18
anupam :0 2014-09-20 16:14 (:0)
anupam pts/0 2014-09-20 16:21 (:0)
$ whoami
anupam
$ who -a system boot 2014-09-20 16:13 run-level 2 2014-09-20 16:13
LOGIN tty4 2014-09-20 16:13 736 id=4
LOGIN tty5 2014-09-20 16:13 740 id=5
anupam - tty2 2014-09-20 16:19 00:01 3200
anupam - tty3 2014-09-20 16:20 . 3346
LOGIN tty6 2014-09-20 16:13 752 id=6
anupam - tty1 2014-09-20 16:18 00:02 3044
anupam ? :0 2014-09-20 16:14 ? 1835 (:0)
anupam + pts/0 2014-09-20 16:21 . 3455 (:0)
$ 

I am not getting some terms in second attribute (- tty2,-tty 3,-tty1i [why - is there in front of them?]) ?:0 (I guess it is indicating my X-window startup [why is there a ? before :0?]), and values at fourth attribute [00:01, ., 00:02, ?, .]?

I tried to look at $ man who -a, but I didn't got these explanation.

1 Answer

• pts/0 is a Pseudo-Terminal Slave (See What does "pts/" in the output of w mean?).

• The (:0) tells you which display you're using.

• the +,-,? tells you whether a user/tty is accepting messages. If true, display a + for each user if mesg y, a - if mesg n, or a ? if their tty cannot be stat'ed.

  See the mesg man page:

  NAME mesg - control write access to your terminal
  SYNOPSIS mesg [y|n]
  DESCRIPTION Mesg controls the access to your terminal by others. It's typically used to allow or disallow other users to write to your terminal (see write(1)).
  OPTIONS y Allow write access to your terminal. n Disallow write access to your terminal. If no option is given, mesg prints out the current access state of your terminal.

Source: who.c