On 2022-08-23 19:15:43 [+0200], Vlastimil Babka wrote: > > +#define slub_local_irq_save(flags) local_irq_save(flags) > > +#define slub_local_irq_restore(flags) local_irq_restore(flags) > > Note these won't be neccessary anymore after > https://lore.kernel.org/linux-mm/20220823170400.26546-6-vbabka@xxxxxxx/T/#u Okay, let me postpone that one and rebase what is left on top. > > @@ -482,7 +488,7 @@ static inline bool __cmpxchg_double_slab(struct kmem_cache *s, struct slab *slab > > void *freelist_new, unsigned long counters_new, > > const char *n) > > { > > - if (!IS_ENABLED(CONFIG_PREEMPT_RT)) > > + if (use_lockless_fast_path()) > > lockdep_assert_irqs_disabled(); > > This test would stay after the patch I referenced above. But while this > change will keep testing the technically correct thing, the name would be > IMHO misleading here, as this is semantically not about the lockless fast > path, but whether we need to have irqs disabled to avoid a deadlock due to > irq incoming when we hold the bit_spin_lock() and its handler trying to > acquire it as well. Color me confused. Memory is never allocated in-IRQ context on PREEMPT_RT. Therefore I don't understand why interrupts must be disabled for the fast path (unless that comment only applied to !RT). It could be about preemption since spinlock, local_lock don't disable preemption and so another allocation on the same CPU is possible. But then you say "we hold the bit_spin_lock()" and this one disables preemption. This means nothing can stop the bit_spin_lock() owner from making progress and since there is no memory allocation in-IRQ, we can't block on the same bit_spin_lock() on the same CPU. Sebastian
