init
 

id:1:initdefault:

init will run the scripts in which directory?

[root@serv01 ~]# ls -l /etc/rc.d
total 112
drwxr-xr-x 2 root root 4096 Mar 24 14:41 init.d
-rwxr-xr-x 1 root root 2255 Sep 21 2006 rc
drwxr-xr-x 2 root root 4096 Mar 24 14:41 rc0.d
drwxr-xr-x 2 root root 4096 Mar 24 14:41 rc1.d
drwxr-xr-x 2 root root 4096 Mar 24 14:41 rc2.d
drwxr-xr-x 2 root root 4096 Mar 24 14:47 rc3.d
drwxr-xr-x 2 root root 4096 Mar 24 14:41 rc4.d
drwxr-xr-x 2 root root 4096 Mar 24 14:47 rc5.d
drwxr-xr-x 2 root root 4096 Mar 24 14:41 rc6.d
-rwxr-xr-x 1 root root 220 Jun 23 2003 rc.local
-rwxr-xr-x 1 root root 26376 Jan 19 2007 rc.sysinit

I am having grave difficulty comprehending init as I thought id:1:initdefault would run state 1 as the default single user mode and run the rc.sysinit script at boot up. Therefore wouldn't it run in the /etc/rc1.d directory?

rc#.d runs in the run level specified. Single user = Run level 1 so would therefore run the items in rc1.d.

More technically it runs the items that begin with S (for start) on entering run level 1 and the items that begin with K (for kill/stop) on exiting run level 1.

Run levels 2 through 5 are "multiuser" run levels that do different things. e.g. Run level 5 is usually the one that starts everything including graphics/X.