>
> Borsenkow Andrej writes:
> > I am sorry if this is stupid question but if it should not work if I
> > understand devfs properly ... I mean these dd's in devfsd.
> >
> > What happens is
> >
> > User accesses /dev/sda4
> > |
> > devfs lookup calls devfsd with LOOKUP
> > |
> > devfsd runs dd if=/dev/sda
> > |
> > |-----------------------+
> > | |
> > devfsd returns to devfs devfs calls devfsd with REGISTER
> > | |
> > devfs repeats lookup devfsd creates /dev/sda4
> > |
> > devfs returns /dev/sda4?
> >
> > So to my eyes there are two independent threads with obvious race
> > condition here. There is no way known to me to synchronize them. It
> > assumes second threads completes BEFORE devfs does second lookup. Is
> > it really garanteed?
>
> If I understand your question: you're wondering about races between
> devfsd and it's child dd process. There are no such races, because
> devfsd will wait for the child process to finish. Devfsd is
> explicitely single-threaded (i.e. it waits for children to finish) to
> avoid any such problems.
>
Not quite. If it is single threaded the thread looks like:
User lookup /dev/sda4 waits
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devfs calls devfsd with LOOKUP /dev/sda4
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devfsd calls dd
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dd returns
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devfsd acknowledges LOOKUP
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devfs repeats lookup /dev/sda4 - but it does not yet exist?
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devfs returns error to user
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devfs calls devfsd with REGISTER .../part4 which is now irrelevant
Please, bear with me. I'd like to be sure I understand how it works and
that it is not just an accident.
-andrej
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