Description of problem:

The grubby(1) man page says, in part:


       There  are  a  number  of  ways  to specify the kernel used for --info,
       --remove-kernel,  and  --update-kernel.  Specificying  DEFAULT  or  ALL
       selects  the  default entry and all of the entries, respectively.  If a
       comma separated list of numbers is given, the boot entries  indexed  by
       those  numbers  are selected. Finally, the title of a boot entry may be
       specified by using TITLE=title as the argument; all entries  with  that
       title are used.

grubby --info=2, for example tells me information about the second entry
in /boot/grub/grub.conf.  But, grubby --remove-kernel=2 results in a grubby
process which does not exit and which uses up lots of cpu time.


Version-Release number of selected component (if applicable):

grubby version 7.0.16

How reproducible: every time

Steps to Reproduce:
1.  Since you probably do not want to clobber your grub.conf file,
    make a copy of it to your home directory.  Here's what I did:

    cp /boot/grub/grub.conf /home/setha/grub.conf

2.  I have seen the problem when running grubby as root, or as my
    normal user setha.  But, become root anyway.  That way you won't
    dismiss this bug because you think it is just a permissions problem.

3.  Here's my grub.conf file from my Fedora 14 installation.  Use this
    or use your own.  I don't think it is specific to this particular
    grub.conf file.  I've also seen the same problem on CentOS with a
    much different grub.conf file.

# grub.conf generated by anaconda
#
# Note that you do not have to rerun grub after making changes to this file
# NOTICE:  You have a /boot partition.  This means that
#          all kernel and initrd paths are relative to /boot/, eg.
#          root (hd0,0)
#          kernel /vmlinuz-version ro root=/dev/sda3
#          initrd /initrd-version.img
# boot=/dev/sda
default=0
timeout=15
# Commented out 3PM 8/12 in an effor to fix the ctl-alt-f2 ctl-alt-f7
# blank screen problem
# splashimage=(hd0,0)/grub/splash.xpm.gz
# hiddenmenu
title Fedora (2.6.35.13-92.fc14.i686)
	root (hd0,0)
	kernel /vmlinuz-2.6.35.13-92.fc14.i686 ro root=LABEL=/ rhgb quiet SYSFONT=latarcyrheb-sun16 LANG=en_US.UTF-8 KEYTABLE=us
	initrd /initramfs-2.6.35.13-92.fc14.i686.img
title Fedora (2.6.35.13-91.fc14.i686)
	root (hd0,0)
	kernel /vmlinuz-2.6.35.13-91.fc14.i686 ro root=LABEL=/ rhgb quiet SYSFONT=latarcyrheb-sun16 LANG=en_US.UTF-8 KEYTABLE=us
	initrd /initramfs-2.6.35.13-91.fc14.i686.img
title Fedora (2.6.35.11-83.fc14.i686)
	root (hd0,0)
	kernel /vmlinuz-2.6.35.11-83.fc14.i686 ro root=LABEL=/ rhgb quiet SYSFONT=latarcyrheb-sun16 LANG=en_US.UTF-8 KEYTABLE=us
	initrd /initramfs-2.6.35.11-83.fc14.i686.img

4.  Try to delete the entry with index 2 from the file.  Since index numbers
start at 0, this should remove the 3rd entry.  Again, I suggest doing this
as root:

    # do this just to show reference by index works:
    grubby --config-file /home/setha/grub.conf --info=2

    # this command does not return
    grubby --config-file /home/setha/grub.conf --remove-kernel=2

  
Actual results:

The --info command returns almost immediately with information about
the entry with index 2 (ie, the 3rd entry) in the grub.conf file:

boot=/dev/sda
index=2
kernel=/vmlinuz-2.6.35.11-83.fc14.i686
args="ro rhgb quiet SYSFONT=latarcyrheb-sun16 LANG=en_US.UTF-8 KEYTABLE=us"
root=LABEL=/
initrd=/boot/initramfs-2.6.35.11-83.fc14.i686.img

The --remove-kernel command runs and runs and runs, eating up cpu time and not actually doing anything.


Expected results:

grubby --remove-kernel=2 would work just like grubby --info=2 works and
just like the man page says it should work.  It will operate on the entry
with index 2 and remove it from the grub.conf file.


Additional info: