Hey all, This topic has been partly discussed here: http://lists.ceph.com/pipermail/ceph-users-ceph.com/2013-March/000799.html Tested on Ceph version 0.67.2. If you create a fresh empty image of, say, 100GB in size on RBD and then use "rbd cp" to make a copy of it, even though the image is sparse, the command will attempt to read every part of it and take far more time than expected. After reading the above thread, I understand why the copy of an essentially empty sparse image on RBD would take so long, but it doesn't explain why the copy won't be sparse itself. If I use "rbd cp" to copy an image, the copy will take it's full allocated size on disk, even if the original was empty. If I use the QEMU "qemu-img"-tool's "convert"-option to convert the original image to the copy without changing the format, essentially only making a copy, it takes it's time as well, but will be faster than "rbd cp" and the resulting copy will be sparse. Example-commands: rbd create --size 102400 test1 rbd cp test1 test2 qemu-img convert -p -f rbd -O rbd rbd:rbd/test1 rbd:rbd/test3 Shouldn't "rbd cp" at least have an option to attempt to sparsify the copy, or copy the sparse parts as sparse? Same goes for "rbd clone", BTW. Regards, Oliver _______________________________________________ ceph-users mailing list ceph-users@xxxxxxxxxxxxxx http://lists.ceph.com/listinfo.cgi/ceph-users-ceph.com
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