Re: [PATCH] mempool: clarify behavior of mempool_alloc_preallocated()

[Vishal Moola (Oracle)]

On Wed, Oct 15, 2025 at 09:27:17PM +0200, Sebastian Andrzej Siewior wrote:
> On 2025-10-15 11:52:24 [-0700], Vishal Moola (Oracle) wrote:
> > > --- a/mm/mempool.c
> > > +++ b/mm/mempool.c
> > > @@ -461,8 +461,8 @@ EXPORT_SYMBOL(mempool_alloc_noprof);
> > >   *             mempool_create().
> > >   *
> > >   * This function is similar to mempool_alloc, but it only attempts allocating
> > > - * an element from the preallocated elements. It does not sleep and immediately
> > > - * returns if no preallocated elements are available.
> > > + * an element from the preallocated elements. It only takes a single spinlock_t
> > 
> > Might it make more sense to say "It may sleep" instead of "takes a
> > single spinlock_t"?
> 
> May sleep usually refers to something that can not be used in an
> interrupt handler.

Gotcha. 

> > I feel like the fact that we take a spinlock isn't the important part
> > here (especially because we always drop it before returning).
> It actually is. A spinlock_t can not be acquired in hardirq context or
> when interrupts are explicitly disabled via local_irq_disable().
> Therefore you should use the function in a local_irq_disable() section.

As someone not too familiar with how the locking intertwines with the
scheduler contexts, seeing something like that makes much more sense
to me than seeing "it only takes a single spinlock_t."