Building from source
You need to start by getting a kernel tree (generally, the more
recent, the better):
host% wget http://www.kernel.org/pub/linux/kernel/v2.6/linux-2.6.24.tar.bz2
Resolving www.kernel.org... 184.108.40.206, 220.127.116.11
Connecting to www.kernel.org|18.104.22.168|:80... connected.
HTTP request sent, awaiting response... 200 OK
Length: 40,845,005 (39M) [application/x-bzip2]
100%[====================================>] 40,845,005 633.10K/s ETA 00:00
15:04:00 (622.07 KB/s) - `linux-2.6.24.tar.bz2' saved [40845005/40845005]
Unpack the tree, which will end up in linux-2.6.24 in this case
host% bunzip2 linux-2.6.24.tar.bz2
host% tar xf linux-2.6.24.tar
host% cd linux-2.6.24
Start with the UML default configuration, which will compile and boot.
If you need to make changes, then do that later using menuconfig or
host% make defconfig ARCH=um
host% # now make menuconfig or xconfig if desired
host% make menuconfig ARCH=um
If you don't start with a defconfig, then the kernel build will be
that of the host (it will find a config file in /boot), which will be
very wrong for UML and will produce a UML that lacks vital drivers and
Note - it is vitally important to put "ARCH=um" on every make command
while building UML, or to "export ARCH=um" to put ARCH in your
environment. This causes the kernel build to build UML, which is a
separate Linux architecture. Not doing so will cause the kernel build
to build or configure a native kernel. If you should forget, clean
the pool like this
host% make mrproper
host% make mrproper ARCH=um
to get rid of all traces of whatever building you did, and start over.
Now, you can start the build
host% make ARCH=um
When this finishes, you will have a UML binary called "linux".
host% ls -l linux
-rwxrwxr-x 2 jdike jdike 18941274 Apr 7 15:18 linux
It's so large because of the debugging symbols built in to it.
Removing those will shrink the UML binary to roughly the size of a
Now, you are ready to boot your new UML.