How do I tell I'm running in a chroot?
I have a unix installation that's supposed to be usable both as a chroot and as a standalone system. If it's running as a chroot, I don't want to run any service (cron, inetd, and so on), because they would conflict with the host system or be redundant.
How do I write a shell script that behaves differently depending on whether it's running in a chroot? My immediate need is a modern Linux system, with /proc
mounted in the chroot, and the script is running as root, but more portable answers are welcome as well. (See How do I tell I'm running in a chroot if /proc is not mounted? for the case of Linux without /proc
.)
More generally, suggestions that work for other containment methods would be interesting. The practical question is, is this system supposed to be running any services? (The answer being no in a chroot, and yes in a full-fledged virtual machines; I don't know about intermediate cases such as jails or containers.)
chroot
add a comment |
I have a unix installation that's supposed to be usable both as a chroot and as a standalone system. If it's running as a chroot, I don't want to run any service (cron, inetd, and so on), because they would conflict with the host system or be redundant.
How do I write a shell script that behaves differently depending on whether it's running in a chroot? My immediate need is a modern Linux system, with /proc
mounted in the chroot, and the script is running as root, but more portable answers are welcome as well. (See How do I tell I'm running in a chroot if /proc is not mounted? for the case of Linux without /proc
.)
More generally, suggestions that work for other containment methods would be interesting. The practical question is, is this system supposed to be running any services? (The answer being no in a chroot, and yes in a full-fledged virtual machines; I don't know about intermediate cases such as jails or containers.)
chroot
add a comment |
I have a unix installation that's supposed to be usable both as a chroot and as a standalone system. If it's running as a chroot, I don't want to run any service (cron, inetd, and so on), because they would conflict with the host system or be redundant.
How do I write a shell script that behaves differently depending on whether it's running in a chroot? My immediate need is a modern Linux system, with /proc
mounted in the chroot, and the script is running as root, but more portable answers are welcome as well. (See How do I tell I'm running in a chroot if /proc is not mounted? for the case of Linux without /proc
.)
More generally, suggestions that work for other containment methods would be interesting. The practical question is, is this system supposed to be running any services? (The answer being no in a chroot, and yes in a full-fledged virtual machines; I don't know about intermediate cases such as jails or containers.)
chroot
I have a unix installation that's supposed to be usable both as a chroot and as a standalone system. If it's running as a chroot, I don't want to run any service (cron, inetd, and so on), because they would conflict with the host system or be redundant.
How do I write a shell script that behaves differently depending on whether it's running in a chroot? My immediate need is a modern Linux system, with /proc
mounted in the chroot, and the script is running as root, but more portable answers are welcome as well. (See How do I tell I'm running in a chroot if /proc is not mounted? for the case of Linux without /proc
.)
More generally, suggestions that work for other containment methods would be interesting. The practical question is, is this system supposed to be running any services? (The answer being no in a chroot, and yes in a full-fledged virtual machines; I don't know about intermediate cases such as jails or containers.)
chroot
chroot
edited Apr 13 '17 at 12:37
Community♦
1
1
asked Jun 3 '11 at 16:55
Gilles
528k12810581583
528k12810581583
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
What I've done here is to test whether the root of the init
process (PID 1) is the same as the root of the current process. Although /proc/1/root
is always a link to /
(unless init
itself is chrooted, but that's not a case I care about), following it leads to the “master” root directory. This technique is used in a few maintenance scripts in Debian, for example to skip starting udev after installation in a chroot.
if [ "$(stat -c %d:%i /)" != "$(stat -c %d:%i /proc/1/root/.)" ]; then
echo "We are chrooted!"
else
echo "Business as usual"
fi
(By the way, this is yet another example of why chroot
is useless for security if the chrooted process has root access. Non-root processes can't read /proc/1/root
, but they can follow /proc/1234/root
if there is a running process with PID 1234 running as the same user.)
If you do not have root permissions, you can look at /proc/1/mountinfo
and /proc/$$/mountinfo
(briefly documented in filesystems/proc.txt
in the Linux kernel documentation). This file is world-readable and contains a lot of information about each mount point in the process's view of the filesystem. The paths in that file are restricted by the chroot affecting the reader process, if any. If the process reading /proc/1/mountinfo
is chrooted into a filesystem that's different from the global root (assuming pid 1's root is the global root), then no entry for /
appears in /proc/1/mountinfo
. If the process reading /proc/1/mountinfo
is chrooted to a directory on the global root filesystem, then an entry for /
appears in /proc/1/mountinfo
, but with a different mount id. Incidentally, the root field ($4
) indicates where the chroot is in its master filesystem.
[ "$(awk '$5=="/" {print $1}' </proc/1/mountinfo)" != "$(awk '$5=="/" {print $1}' </proc/$$/mountinfo)" ]
This is a pure Linux solution. It may be generalizable to other Unix variants with a sufficiently similar /proc
(Solaris has a similar /proc/1/root
, I think, but not mountinfo
).
1
This won't work in OpenBSD because it has random PIDs; the root process is basically never PID 1. Now you know why!
– Adam Katz
Jan 16 '15 at 5:59
@AdamKatz "... with a couple of obvious exceptions, e.g., init(8)." So which is it?
– muru
Jan 16 '15 at 9:40
@muru: aw, shucks. You've shot me down. I'm not sure whyinit(8)
would absolutely need to have the #1 slot unless there's some kind of hard-coded nature that requires it (in which I'd still be unsure as to why). Of course, the BSDs have much more advanced jails than just chroot, so I'm not even sure how problematic this is.
– Adam Katz
Jan 16 '15 at 10:02
3
@AdamKatz It's the opposite: pid 1 has a special role (it must reap zombies, and it is immune to SIGKILL). The init program is an implementation of that role. The reason my answer doesn't work in OpenBSD has nothing to do with this: it's because OpenBSD doesn't have anything like Solaris/Linux's/proc
. My answer wasn't meant to address anything but Linux anyway.
– Gilles
Jan 16 '15 at 11:40
@Gilles I figured OpenBSD would have this defeated in some manner or other. Still, I'm surprised that all of those special role items aren't capable of being applied to an arbitrary PID (without consequences), which is what I meant in my italicized "why" earlier.
– Adam Katz
Jan 16 '15 at 17:11
|
show 6 more comments
As mentioned in Portable way to find inode number and Detecting a chroot jail from within, you can check whether the inode number of /
is 2
:
$ ls -di /
2 /
An inode number that's different from 2 indicates that the apparent root is not the actual root of a filesystem. This will not detect chroots that happen to be rooted on a mount point, or on operating systems with random root inode numbers.
On what filesystems does this heuristic work?
– Gilles
Nov 9 '11 at 11:36
Tested on ext3 and hfs.
– l0b0
Nov 9 '11 at 11:52
So I was fooling around, and I think I've found a more reliable method that doesn't require root permissions (Linux only). I'm still open to counter-examples or more portable methods.
– Gilles
Nov 9 '11 at 19:15
6
This is true for ext[234], but not of all filesystems. It also only tests that your root is the root of the filesystem, which may not be mounted as the real root. In other words, if you mount another partition in /jail andchroot /jail
, then it will look like the real root to this test.
– psusi
Nov 9 '11 at 19:26
1
@AdamKatz Apparently not. Tested in openbsd 6.0-stable, the inode number is still 2 for the actual root path while it's a random number for the chroot.
– Dmitri DB
Sep 16 '16 at 3:11
|
show 4 more comments
While clearly not as portable as many other options listed here, if you're on a Debian-based system, try ischroot
.
See: https://manpages.debian.org/jessie/debianutils/ischroot.1.en.html
To get the status in the console directly, using ischroot:
ischroot;echo $?
Exit codes:
0 if currently running in a chroot
1 if currently not running in a chroot
2 if the detection is not possible (On GNU/Linux this happens if the script is not run as root).
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
What I've done here is to test whether the root of the init
process (PID 1) is the same as the root of the current process. Although /proc/1/root
is always a link to /
(unless init
itself is chrooted, but that's not a case I care about), following it leads to the “master” root directory. This technique is used in a few maintenance scripts in Debian, for example to skip starting udev after installation in a chroot.
if [ "$(stat -c %d:%i /)" != "$(stat -c %d:%i /proc/1/root/.)" ]; then
echo "We are chrooted!"
else
echo "Business as usual"
fi
(By the way, this is yet another example of why chroot
is useless for security if the chrooted process has root access. Non-root processes can't read /proc/1/root
, but they can follow /proc/1234/root
if there is a running process with PID 1234 running as the same user.)
If you do not have root permissions, you can look at /proc/1/mountinfo
and /proc/$$/mountinfo
(briefly documented in filesystems/proc.txt
in the Linux kernel documentation). This file is world-readable and contains a lot of information about each mount point in the process's view of the filesystem. The paths in that file are restricted by the chroot affecting the reader process, if any. If the process reading /proc/1/mountinfo
is chrooted into a filesystem that's different from the global root (assuming pid 1's root is the global root), then no entry for /
appears in /proc/1/mountinfo
. If the process reading /proc/1/mountinfo
is chrooted to a directory on the global root filesystem, then an entry for /
appears in /proc/1/mountinfo
, but with a different mount id. Incidentally, the root field ($4
) indicates where the chroot is in its master filesystem.
[ "$(awk '$5=="/" {print $1}' </proc/1/mountinfo)" != "$(awk '$5=="/" {print $1}' </proc/$$/mountinfo)" ]
This is a pure Linux solution. It may be generalizable to other Unix variants with a sufficiently similar /proc
(Solaris has a similar /proc/1/root
, I think, but not mountinfo
).
1
This won't work in OpenBSD because it has random PIDs; the root process is basically never PID 1. Now you know why!
– Adam Katz
Jan 16 '15 at 5:59
@AdamKatz "... with a couple of obvious exceptions, e.g., init(8)." So which is it?
– muru
Jan 16 '15 at 9:40
@muru: aw, shucks. You've shot me down. I'm not sure whyinit(8)
would absolutely need to have the #1 slot unless there's some kind of hard-coded nature that requires it (in which I'd still be unsure as to why). Of course, the BSDs have much more advanced jails than just chroot, so I'm not even sure how problematic this is.
– Adam Katz
Jan 16 '15 at 10:02
3
@AdamKatz It's the opposite: pid 1 has a special role (it must reap zombies, and it is immune to SIGKILL). The init program is an implementation of that role. The reason my answer doesn't work in OpenBSD has nothing to do with this: it's because OpenBSD doesn't have anything like Solaris/Linux's/proc
. My answer wasn't meant to address anything but Linux anyway.
– Gilles
Jan 16 '15 at 11:40
@Gilles I figured OpenBSD would have this defeated in some manner or other. Still, I'm surprised that all of those special role items aren't capable of being applied to an arbitrary PID (without consequences), which is what I meant in my italicized "why" earlier.
– Adam Katz
Jan 16 '15 at 17:11
|
show 6 more comments
What I've done here is to test whether the root of the init
process (PID 1) is the same as the root of the current process. Although /proc/1/root
is always a link to /
(unless init
itself is chrooted, but that's not a case I care about), following it leads to the “master” root directory. This technique is used in a few maintenance scripts in Debian, for example to skip starting udev after installation in a chroot.
if [ "$(stat -c %d:%i /)" != "$(stat -c %d:%i /proc/1/root/.)" ]; then
echo "We are chrooted!"
else
echo "Business as usual"
fi
(By the way, this is yet another example of why chroot
is useless for security if the chrooted process has root access. Non-root processes can't read /proc/1/root
, but they can follow /proc/1234/root
if there is a running process with PID 1234 running as the same user.)
If you do not have root permissions, you can look at /proc/1/mountinfo
and /proc/$$/mountinfo
(briefly documented in filesystems/proc.txt
in the Linux kernel documentation). This file is world-readable and contains a lot of information about each mount point in the process's view of the filesystem. The paths in that file are restricted by the chroot affecting the reader process, if any. If the process reading /proc/1/mountinfo
is chrooted into a filesystem that's different from the global root (assuming pid 1's root is the global root), then no entry for /
appears in /proc/1/mountinfo
. If the process reading /proc/1/mountinfo
is chrooted to a directory on the global root filesystem, then an entry for /
appears in /proc/1/mountinfo
, but with a different mount id. Incidentally, the root field ($4
) indicates where the chroot is in its master filesystem.
[ "$(awk '$5=="/" {print $1}' </proc/1/mountinfo)" != "$(awk '$5=="/" {print $1}' </proc/$$/mountinfo)" ]
This is a pure Linux solution. It may be generalizable to other Unix variants with a sufficiently similar /proc
(Solaris has a similar /proc/1/root
, I think, but not mountinfo
).
1
This won't work in OpenBSD because it has random PIDs; the root process is basically never PID 1. Now you know why!
– Adam Katz
Jan 16 '15 at 5:59
@AdamKatz "... with a couple of obvious exceptions, e.g., init(8)." So which is it?
– muru
Jan 16 '15 at 9:40
@muru: aw, shucks. You've shot me down. I'm not sure whyinit(8)
would absolutely need to have the #1 slot unless there's some kind of hard-coded nature that requires it (in which I'd still be unsure as to why). Of course, the BSDs have much more advanced jails than just chroot, so I'm not even sure how problematic this is.
– Adam Katz
Jan 16 '15 at 10:02
3
@AdamKatz It's the opposite: pid 1 has a special role (it must reap zombies, and it is immune to SIGKILL). The init program is an implementation of that role. The reason my answer doesn't work in OpenBSD has nothing to do with this: it's because OpenBSD doesn't have anything like Solaris/Linux's/proc
. My answer wasn't meant to address anything but Linux anyway.
– Gilles
Jan 16 '15 at 11:40
@Gilles I figured OpenBSD would have this defeated in some manner or other. Still, I'm surprised that all of those special role items aren't capable of being applied to an arbitrary PID (without consequences), which is what I meant in my italicized "why" earlier.
– Adam Katz
Jan 16 '15 at 17:11
|
show 6 more comments
What I've done here is to test whether the root of the init
process (PID 1) is the same as the root of the current process. Although /proc/1/root
is always a link to /
(unless init
itself is chrooted, but that's not a case I care about), following it leads to the “master” root directory. This technique is used in a few maintenance scripts in Debian, for example to skip starting udev after installation in a chroot.
if [ "$(stat -c %d:%i /)" != "$(stat -c %d:%i /proc/1/root/.)" ]; then
echo "We are chrooted!"
else
echo "Business as usual"
fi
(By the way, this is yet another example of why chroot
is useless for security if the chrooted process has root access. Non-root processes can't read /proc/1/root
, but they can follow /proc/1234/root
if there is a running process with PID 1234 running as the same user.)
If you do not have root permissions, you can look at /proc/1/mountinfo
and /proc/$$/mountinfo
(briefly documented in filesystems/proc.txt
in the Linux kernel documentation). This file is world-readable and contains a lot of information about each mount point in the process's view of the filesystem. The paths in that file are restricted by the chroot affecting the reader process, if any. If the process reading /proc/1/mountinfo
is chrooted into a filesystem that's different from the global root (assuming pid 1's root is the global root), then no entry for /
appears in /proc/1/mountinfo
. If the process reading /proc/1/mountinfo
is chrooted to a directory on the global root filesystem, then an entry for /
appears in /proc/1/mountinfo
, but with a different mount id. Incidentally, the root field ($4
) indicates where the chroot is in its master filesystem.
[ "$(awk '$5=="/" {print $1}' </proc/1/mountinfo)" != "$(awk '$5=="/" {print $1}' </proc/$$/mountinfo)" ]
This is a pure Linux solution. It may be generalizable to other Unix variants with a sufficiently similar /proc
(Solaris has a similar /proc/1/root
, I think, but not mountinfo
).
What I've done here is to test whether the root of the init
process (PID 1) is the same as the root of the current process. Although /proc/1/root
is always a link to /
(unless init
itself is chrooted, but that's not a case I care about), following it leads to the “master” root directory. This technique is used in a few maintenance scripts in Debian, for example to skip starting udev after installation in a chroot.
if [ "$(stat -c %d:%i /)" != "$(stat -c %d:%i /proc/1/root/.)" ]; then
echo "We are chrooted!"
else
echo "Business as usual"
fi
(By the way, this is yet another example of why chroot
is useless for security if the chrooted process has root access. Non-root processes can't read /proc/1/root
, but they can follow /proc/1234/root
if there is a running process with PID 1234 running as the same user.)
If you do not have root permissions, you can look at /proc/1/mountinfo
and /proc/$$/mountinfo
(briefly documented in filesystems/proc.txt
in the Linux kernel documentation). This file is world-readable and contains a lot of information about each mount point in the process's view of the filesystem. The paths in that file are restricted by the chroot affecting the reader process, if any. If the process reading /proc/1/mountinfo
is chrooted into a filesystem that's different from the global root (assuming pid 1's root is the global root), then no entry for /
appears in /proc/1/mountinfo
. If the process reading /proc/1/mountinfo
is chrooted to a directory on the global root filesystem, then an entry for /
appears in /proc/1/mountinfo
, but with a different mount id. Incidentally, the root field ($4
) indicates where the chroot is in its master filesystem.
[ "$(awk '$5=="/" {print $1}' </proc/1/mountinfo)" != "$(awk '$5=="/" {print $1}' </proc/$$/mountinfo)" ]
This is a pure Linux solution. It may be generalizable to other Unix variants with a sufficiently similar /proc
(Solaris has a similar /proc/1/root
, I think, but not mountinfo
).
edited Jan 16 '15 at 11:43
answered Jun 3 '11 at 16:55
Gilles
528k12810581583
528k12810581583
1
This won't work in OpenBSD because it has random PIDs; the root process is basically never PID 1. Now you know why!
– Adam Katz
Jan 16 '15 at 5:59
@AdamKatz "... with a couple of obvious exceptions, e.g., init(8)." So which is it?
– muru
Jan 16 '15 at 9:40
@muru: aw, shucks. You've shot me down. I'm not sure whyinit(8)
would absolutely need to have the #1 slot unless there's some kind of hard-coded nature that requires it (in which I'd still be unsure as to why). Of course, the BSDs have much more advanced jails than just chroot, so I'm not even sure how problematic this is.
– Adam Katz
Jan 16 '15 at 10:02
3
@AdamKatz It's the opposite: pid 1 has a special role (it must reap zombies, and it is immune to SIGKILL). The init program is an implementation of that role. The reason my answer doesn't work in OpenBSD has nothing to do with this: it's because OpenBSD doesn't have anything like Solaris/Linux's/proc
. My answer wasn't meant to address anything but Linux anyway.
– Gilles
Jan 16 '15 at 11:40
@Gilles I figured OpenBSD would have this defeated in some manner or other. Still, I'm surprised that all of those special role items aren't capable of being applied to an arbitrary PID (without consequences), which is what I meant in my italicized "why" earlier.
– Adam Katz
Jan 16 '15 at 17:11
|
show 6 more comments
1
This won't work in OpenBSD because it has random PIDs; the root process is basically never PID 1. Now you know why!
– Adam Katz
Jan 16 '15 at 5:59
@AdamKatz "... with a couple of obvious exceptions, e.g., init(8)." So which is it?
– muru
Jan 16 '15 at 9:40
@muru: aw, shucks. You've shot me down. I'm not sure whyinit(8)
would absolutely need to have the #1 slot unless there's some kind of hard-coded nature that requires it (in which I'd still be unsure as to why). Of course, the BSDs have much more advanced jails than just chroot, so I'm not even sure how problematic this is.
– Adam Katz
Jan 16 '15 at 10:02
3
@AdamKatz It's the opposite: pid 1 has a special role (it must reap zombies, and it is immune to SIGKILL). The init program is an implementation of that role. The reason my answer doesn't work in OpenBSD has nothing to do with this: it's because OpenBSD doesn't have anything like Solaris/Linux's/proc
. My answer wasn't meant to address anything but Linux anyway.
– Gilles
Jan 16 '15 at 11:40
@Gilles I figured OpenBSD would have this defeated in some manner or other. Still, I'm surprised that all of those special role items aren't capable of being applied to an arbitrary PID (without consequences), which is what I meant in my italicized "why" earlier.
– Adam Katz
Jan 16 '15 at 17:11
1
1
This won't work in OpenBSD because it has random PIDs; the root process is basically never PID 1. Now you know why!
– Adam Katz
Jan 16 '15 at 5:59
This won't work in OpenBSD because it has random PIDs; the root process is basically never PID 1. Now you know why!
– Adam Katz
Jan 16 '15 at 5:59
@AdamKatz "... with a couple of obvious exceptions, e.g., init(8)." So which is it?
– muru
Jan 16 '15 at 9:40
@AdamKatz "... with a couple of obvious exceptions, e.g., init(8)." So which is it?
– muru
Jan 16 '15 at 9:40
@muru: aw, shucks. You've shot me down. I'm not sure why
init(8)
would absolutely need to have the #1 slot unless there's some kind of hard-coded nature that requires it (in which I'd still be unsure as to why). Of course, the BSDs have much more advanced jails than just chroot, so I'm not even sure how problematic this is.– Adam Katz
Jan 16 '15 at 10:02
@muru: aw, shucks. You've shot me down. I'm not sure why
init(8)
would absolutely need to have the #1 slot unless there's some kind of hard-coded nature that requires it (in which I'd still be unsure as to why). Of course, the BSDs have much more advanced jails than just chroot, so I'm not even sure how problematic this is.– Adam Katz
Jan 16 '15 at 10:02
3
3
@AdamKatz It's the opposite: pid 1 has a special role (it must reap zombies, and it is immune to SIGKILL). The init program is an implementation of that role. The reason my answer doesn't work in OpenBSD has nothing to do with this: it's because OpenBSD doesn't have anything like Solaris/Linux's
/proc
. My answer wasn't meant to address anything but Linux anyway.– Gilles
Jan 16 '15 at 11:40
@AdamKatz It's the opposite: pid 1 has a special role (it must reap zombies, and it is immune to SIGKILL). The init program is an implementation of that role. The reason my answer doesn't work in OpenBSD has nothing to do with this: it's because OpenBSD doesn't have anything like Solaris/Linux's
/proc
. My answer wasn't meant to address anything but Linux anyway.– Gilles
Jan 16 '15 at 11:40
@Gilles I figured OpenBSD would have this defeated in some manner or other. Still, I'm surprised that all of those special role items aren't capable of being applied to an arbitrary PID (without consequences), which is what I meant in my italicized "why" earlier.
– Adam Katz
Jan 16 '15 at 17:11
@Gilles I figured OpenBSD would have this defeated in some manner or other. Still, I'm surprised that all of those special role items aren't capable of being applied to an arbitrary PID (without consequences), which is what I meant in my italicized "why" earlier.
– Adam Katz
Jan 16 '15 at 17:11
|
show 6 more comments
As mentioned in Portable way to find inode number and Detecting a chroot jail from within, you can check whether the inode number of /
is 2
:
$ ls -di /
2 /
An inode number that's different from 2 indicates that the apparent root is not the actual root of a filesystem. This will not detect chroots that happen to be rooted on a mount point, or on operating systems with random root inode numbers.
On what filesystems does this heuristic work?
– Gilles
Nov 9 '11 at 11:36
Tested on ext3 and hfs.
– l0b0
Nov 9 '11 at 11:52
So I was fooling around, and I think I've found a more reliable method that doesn't require root permissions (Linux only). I'm still open to counter-examples or more portable methods.
– Gilles
Nov 9 '11 at 19:15
6
This is true for ext[234], but not of all filesystems. It also only tests that your root is the root of the filesystem, which may not be mounted as the real root. In other words, if you mount another partition in /jail andchroot /jail
, then it will look like the real root to this test.
– psusi
Nov 9 '11 at 19:26
1
@AdamKatz Apparently not. Tested in openbsd 6.0-stable, the inode number is still 2 for the actual root path while it's a random number for the chroot.
– Dmitri DB
Sep 16 '16 at 3:11
|
show 4 more comments
As mentioned in Portable way to find inode number and Detecting a chroot jail from within, you can check whether the inode number of /
is 2
:
$ ls -di /
2 /
An inode number that's different from 2 indicates that the apparent root is not the actual root of a filesystem. This will not detect chroots that happen to be rooted on a mount point, or on operating systems with random root inode numbers.
On what filesystems does this heuristic work?
– Gilles
Nov 9 '11 at 11:36
Tested on ext3 and hfs.
– l0b0
Nov 9 '11 at 11:52
So I was fooling around, and I think I've found a more reliable method that doesn't require root permissions (Linux only). I'm still open to counter-examples or more portable methods.
– Gilles
Nov 9 '11 at 19:15
6
This is true for ext[234], but not of all filesystems. It also only tests that your root is the root of the filesystem, which may not be mounted as the real root. In other words, if you mount another partition in /jail andchroot /jail
, then it will look like the real root to this test.
– psusi
Nov 9 '11 at 19:26
1
@AdamKatz Apparently not. Tested in openbsd 6.0-stable, the inode number is still 2 for the actual root path while it's a random number for the chroot.
– Dmitri DB
Sep 16 '16 at 3:11
|
show 4 more comments
As mentioned in Portable way to find inode number and Detecting a chroot jail from within, you can check whether the inode number of /
is 2
:
$ ls -di /
2 /
An inode number that's different from 2 indicates that the apparent root is not the actual root of a filesystem. This will not detect chroots that happen to be rooted on a mount point, or on operating systems with random root inode numbers.
As mentioned in Portable way to find inode number and Detecting a chroot jail from within, you can check whether the inode number of /
is 2
:
$ ls -di /
2 /
An inode number that's different from 2 indicates that the apparent root is not the actual root of a filesystem. This will not detect chroots that happen to be rooted on a mount point, or on operating systems with random root inode numbers.
edited May 23 '17 at 12:39
Community♦
1
1
answered Nov 9 '11 at 8:24
l0b0
27.6k17113242
27.6k17113242
On what filesystems does this heuristic work?
– Gilles
Nov 9 '11 at 11:36
Tested on ext3 and hfs.
– l0b0
Nov 9 '11 at 11:52
So I was fooling around, and I think I've found a more reliable method that doesn't require root permissions (Linux only). I'm still open to counter-examples or more portable methods.
– Gilles
Nov 9 '11 at 19:15
6
This is true for ext[234], but not of all filesystems. It also only tests that your root is the root of the filesystem, which may not be mounted as the real root. In other words, if you mount another partition in /jail andchroot /jail
, then it will look like the real root to this test.
– psusi
Nov 9 '11 at 19:26
1
@AdamKatz Apparently not. Tested in openbsd 6.0-stable, the inode number is still 2 for the actual root path while it's a random number for the chroot.
– Dmitri DB
Sep 16 '16 at 3:11
|
show 4 more comments
On what filesystems does this heuristic work?
– Gilles
Nov 9 '11 at 11:36
Tested on ext3 and hfs.
– l0b0
Nov 9 '11 at 11:52
So I was fooling around, and I think I've found a more reliable method that doesn't require root permissions (Linux only). I'm still open to counter-examples or more portable methods.
– Gilles
Nov 9 '11 at 19:15
6
This is true for ext[234], but not of all filesystems. It also only tests that your root is the root of the filesystem, which may not be mounted as the real root. In other words, if you mount another partition in /jail andchroot /jail
, then it will look like the real root to this test.
– psusi
Nov 9 '11 at 19:26
1
@AdamKatz Apparently not. Tested in openbsd 6.0-stable, the inode number is still 2 for the actual root path while it's a random number for the chroot.
– Dmitri DB
Sep 16 '16 at 3:11
On what filesystems does this heuristic work?
– Gilles
Nov 9 '11 at 11:36
On what filesystems does this heuristic work?
– Gilles
Nov 9 '11 at 11:36
Tested on ext3 and hfs.
– l0b0
Nov 9 '11 at 11:52
Tested on ext3 and hfs.
– l0b0
Nov 9 '11 at 11:52
So I was fooling around, and I think I've found a more reliable method that doesn't require root permissions (Linux only). I'm still open to counter-examples or more portable methods.
– Gilles
Nov 9 '11 at 19:15
So I was fooling around, and I think I've found a more reliable method that doesn't require root permissions (Linux only). I'm still open to counter-examples or more portable methods.
– Gilles
Nov 9 '11 at 19:15
6
6
This is true for ext[234], but not of all filesystems. It also only tests that your root is the root of the filesystem, which may not be mounted as the real root. In other words, if you mount another partition in /jail and
chroot /jail
, then it will look like the real root to this test.– psusi
Nov 9 '11 at 19:26
This is true for ext[234], but not of all filesystems. It also only tests that your root is the root of the filesystem, which may not be mounted as the real root. In other words, if you mount another partition in /jail and
chroot /jail
, then it will look like the real root to this test.– psusi
Nov 9 '11 at 19:26
1
1
@AdamKatz Apparently not. Tested in openbsd 6.0-stable, the inode number is still 2 for the actual root path while it's a random number for the chroot.
– Dmitri DB
Sep 16 '16 at 3:11
@AdamKatz Apparently not. Tested in openbsd 6.0-stable, the inode number is still 2 for the actual root path while it's a random number for the chroot.
– Dmitri DB
Sep 16 '16 at 3:11
|
show 4 more comments
While clearly not as portable as many other options listed here, if you're on a Debian-based system, try ischroot
.
See: https://manpages.debian.org/jessie/debianutils/ischroot.1.en.html
To get the status in the console directly, using ischroot:
ischroot;echo $?
Exit codes:
0 if currently running in a chroot
1 if currently not running in a chroot
2 if the detection is not possible (On GNU/Linux this happens if the script is not run as root).
add a comment |
While clearly not as portable as many other options listed here, if you're on a Debian-based system, try ischroot
.
See: https://manpages.debian.org/jessie/debianutils/ischroot.1.en.html
To get the status in the console directly, using ischroot:
ischroot;echo $?
Exit codes:
0 if currently running in a chroot
1 if currently not running in a chroot
2 if the detection is not possible (On GNU/Linux this happens if the script is not run as root).
add a comment |
While clearly not as portable as many other options listed here, if you're on a Debian-based system, try ischroot
.
See: https://manpages.debian.org/jessie/debianutils/ischroot.1.en.html
To get the status in the console directly, using ischroot:
ischroot;echo $?
Exit codes:
0 if currently running in a chroot
1 if currently not running in a chroot
2 if the detection is not possible (On GNU/Linux this happens if the script is not run as root).
While clearly not as portable as many other options listed here, if you're on a Debian-based system, try ischroot
.
See: https://manpages.debian.org/jessie/debianutils/ischroot.1.en.html
To get the status in the console directly, using ischroot:
ischroot;echo $?
Exit codes:
0 if currently running in a chroot
1 if currently not running in a chroot
2 if the detection is not possible (On GNU/Linux this happens if the script is not run as root).
edited Dec 17 at 23:43
terdon♦
128k31249423
128k31249423
answered Jan 10 at 15:18
thom_nic
17810
17810
add a comment |
add a comment |
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