On 28Mar2019 01:12, Paulo da Silva <p_s_d_a_s_i_l_v_a...@netcabo.pt> wrote:
Às 23:09 de 27/03/19, Cameron Simpson escreveu:
On 27Mar2019 21:49, Paulo da Silva <p_s_d_a_s_i_l_v_a...@netcabo.pt> wrote:
...
The filefrag manual entry says it works by calling one of 2 ioctls. You
can do that from Python with the ioctl() function in the standard fcntl
module. I haven't tried to do this, but the results should be basicly as
fast as filefrag itself.
You'll need to decode the result the ioctl returns of course.
Thanks Cameron, I'll take a look at that.
Oh, just tangential to this.
If you were doing this ad hoc, yes calling the filefrag executable is
very expensive. But if you are always doing a large batch of filenames
invoking:
filefrag lots of filenames here ...
and reading from its output can be very effective, because the expense
of the executable is amortized over all the files - the per file cost is
much reduced. And it saves you working out how to use the ioctls from
Python :-)
You'll learn more from going the ioctl route though, and it gives you
complete control if you need it.
Cheers,
Cameron Simpson <c...@cskk.id.au>
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