On Sat, Apr 30, 2022 at 12:31:07PM -0000, Curt wrote: > On 2022-04-30, Thomas Schmitt <scdbackup@gmx.net> wrote: > > > > Indeed. With normal filesystem operations there should be no need to call > > something like sync(2) in order to get a consistent representation of the > > current filesystem state. > > > > What does the following mean, then, in that light: > > Because of delayed allocation and other performance optimizations, ext4's > behavior of writing files to disk is different from ext3. In ext4, when a > program writes to the file system, it is not guaranteed to be on-disk unless > the program issues an fsync() call afterwards. You still can't observe that during a normally running system. You'll see the same file system behaviour regardless of whether the data have touched the disk or are still in cache. The only way to actually observe this in action is to "pull the plug" before data has a chance to reach the disk. Then, of course, some files you thought were there will have magically disappeared. Cheers -- t
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