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,-tty1
i [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.
Solution 1:
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 ifmesg y
, a-
ifmesg 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