UNIX for Advanced & Expert Users

Hi,

I have a query regarding /dev/loop0 device.

Let's say that I execute the following steps

Code:

root@ubuntu:/tmp# dd if=/dev/zero of=test.img bs=1024 count=300000
root@ubuntu:/tmp# losetup /dev/loop0 test.img
root@ubuntu:/tmp# mkfs -t ext3 /dev/loop0
root@ubuntu:/tmp# mount /dev/loop0 /tmp/mnt

Now, if I create & write into a file /tmp/mnt/fileOne,
will there be a any overhead compared to writing a file directly on /tmp/fileOne?

The difference is that, in the first case I am writing into the /dev/loop0 device. I understand that the /dev/loop0 will have its own inode table and other entries required to retrieve/access the file.

But would there be any difference related to read/write buffers allocated to the device.

Will there be two levels of indirection in this case?
(i,e) A separate RW buffer for /dev/loop0 taken care of by the loop device driver and another level of RW buffer for the physical harddrive?

Any pointers in this regard would be useful.

Thanks,
Alphonse

I'd expect extra overhead on open, only. Once open, the fd writes to a disk and updates an inode on disk, either way. There might be more contention for any device level locks for /tmp device than the new, relatively exclusive device.

Try some io with and without fsync, close-open, seeks (to flush buffers) and see.

There hasn't been any reason to fiddle with losetup by hand for a very long time. mount can, and should, handle loop devices itself.

Code:

mount -o loop file.img directory/