Is it possible to create a directory and file inside /proc/sys?

Question

I am running CentOS 6.2 and I need to create a subdirectory named "crypto" inside /proc/sys. Inside /proc/sys/crypto, I need to create a file named test which contains the value "1".

Answer 1

Ignacio Vazquez-Abrams's answer is correct: files under /proc and /sys are provided by the kernel. If you want to add a file there, write a kernel module. You can find a slightly dated presentation of the /proc programming interface in Linux Device Drivers 3rd ed. ch. 4.

Technically, it is possible to jump through hoops to make a file appear in /proc/sys/crypto (or anywhere else, really). Make a bind mount from /proc/sys/crypto to a staging directory, and make a union mount of the staging directory and an overlay directory back onto /proc/sys/crypto. Here's an example using unionfs-fuse.

# mkdir /tmp/original /tmp/overlay
# mount --bind /proc/sys/crypto /tmp/original
# unionfs-fuse -o nonempty /tmp/overlay=RW:/tmp/original=RO /proc/sys/crypto
# echo hello >/proc/sys/crypto/test
# cat /proc/sys/crypto/test
hello
# umount /proc/sys/crypto
# umount /proc/sys/crypto
# cat /tmp/overlay/test
hello

Note: I disclaim any responsibility for system or brain damage caused by experimenting with this stuff. The commands above are perfectly safe, but messing with other areas of /sys and /proc can cause weird behavior.

Answer 2

/proc is a virtual filesystem. You will need to write a kernel module that creates the appropriate structures within it.

Answer 3

You probably don't want to do this, but if you do need to create a directory in /proc this is all you need in a kernel module

#include <linux/module.h>
#include <linux/proc_fs.h>

static int procfoo_init(void) {
    proc_mkdir("foo", NULL);
    return 0;
}

module_init(procfoo_init);

Once you have a mount point in /proc you can bind a normal fs path to it and do what you like in the directory from userland.

mount --bind /var/lib/foo /proc/foo

This will break as soon as you upgrade the kernel, and almost certainly not what op requires, but here it is since this comes up in search engines.