Re: initialize a mutex into locked state?

[Peter Zijlstra]

On Fri, Jun 17, 2016 at 10:24:32AM -0400, Oleg Drokin wrote:
> 
> On Jun 17, 2016, at 10:19 AM, Peter Zijlstra wrote:
> 
> > On Fri, Jun 17, 2016 at 10:14:10AM -0400, Oleg Drokin wrote:
> >> 
> >> On Jun 17, 2016, at 4:25 AM, Peter Zijlstra wrote:
> >> 
> >>> On Wed, Jun 15, 2016 at 02:23:35PM -0400, Oleg Drokin wrote:
> >>>> Hello!
> >>>> 
> >>>> To my surprise I found out that it's not possible to initialise a mutex into
> >>>> a locked state.
> >>>> I discussed it with Arjan and apparently there's no fundamental reason
> >>>> not to allow this.
> >>> 
> >>> There is. A mutex _must_ have an owner. If you can initialize it in
> >>> locked state, you could do so statically, ie. outside of the context of
> >>> a task.
> >> 
> >> What's wrong with disallowing only static initializers, but allowing dynamic ones?
> >> Then there is a clear owner.
> > 
> > At which point, what wrong with the simple:
> > 
> > 	mutex_init(&m);
> > 	mutex_lock(&m);
> > 
> > Sequence? Its obvious, has clear semantics and doesn't extend the API.
> 
> The problem is:
> 
> spin_lock(somelock);
> structure = some_internal_list_lookup(list);
> if (structure)
> 	goto out;
> 
> init_new_structure(new_structure);
> mutex_init(&new_structure->s_mutex);
> mutex_lock(&new_structure->s_mutex);  // XXX CANNOT DO THIS UNDER SPINLOCK!

	mutex_trylock(&new_structure->s_mutex);

should work, since you know it cannot be acquired yet by anybody else,
since you've not published it yet.

And a trylock does not sleep, so is perfectly fine under a spinlock.

> 
> list_add(list, new_structure->s_list);
> structure = new_structure;
> out:
> spin_unlock(somelock);
> return structure;
>