Re: [RFC] Raceless module interface

[Daniel Phillips <phillips@arcor.de>]

On Friday 13 September 2002 16:59, Thunder from the hill wrote:
> Hi,
> 
> On Fri, 13 Sep 2002, Daniel Phillips wrote:
> > On Friday 13 September 2002 16:33, Thunder from the hill wrote:
> > > Look, first we watch the module initialization, that is, we run the 
> > > critical stuff like resource allocation, data structure allocation, etc. 
> > > If we fail here, we can't load the module, because it would be unoperative 
> > > if we proceed. (Because the data simply isn't there.)
> 
> -> if we don't do the starting here, we can operate on the data structures
>    earlier, since we know they're running free.
> 
> Also could we run start again, even though it sounds buggy.

We can do this without having start separte from init as well.

> > please identify the race Rusty avoided and show how I did not avoid the
> > same race.
> 
> I'm sure Rusty could do that better.

I'd be surprised if Rusty can do it any better than you.  It's hard to
show a race that doesn't exist, even harder to prove that a four-prong
interface is necessary in order to be able to handle it.  The latter is
the question on the table.

> However, there might be some weird 
> situations. For example, take someone trying to bring all modules down 
> the moment we init. We might start running in unchecked environment, and 
> there we fail because there is no 'we' any more.

Oh indeed, there are weird situations, but they apply equally to the
two-prong and the four-prong interfaces.

> Thus rather module->init(). if (module) module->start(). Since then we can 
> be sure that the module is locked, and if somebody unloads it, he'll have 
> to wait for the use count to drop.

This applies equally to the two-prong interface.

> Or as another example, take someone trying to use the resources we claimed
> before the module is really up. If you can rely on the module to be known
> to be up, you know what do do. Yes, usually that's no real good example, 
> since resources ought to be locked as well.

This applies equally to the two-prong interface.  Do you see the pattern
yet?

-- 
Daniel