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