Re: [PATCH] mm/page_alloc: Use write_seqlock_irqsave() instead write_seqlock() + local_irq_save().

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On 2023/06/21 23:50, Tetsuo Handa wrote:
> On 2023/06/21 23:34, Sebastian Andrzej Siewior wrote:
>>> Also, if local_irq_save() is hidden due to RT, what guarantees that
>>>
>>>   write_seqlock_irqsave(&zonelist_update_seq, flags);
>>>   <<IRQ>>
>>>     some_timer_function() {
>>>       printk();
>>>     }
>>>   <<IRQ>>
>>>   printk_deferred_enter();
>>>
>>> does not happen because write_seqlock_irqsave() does not disable IRQ?
>>
>> I don't see how zonelist_update_seq and printk here are connected
>> without the port lock/ or memory allocation. But there are two things
>> that are different on RT which probably answer your question:
> 
> It is explained as the first deadlock scenario in commit 1007843a9190
> ("mm/page_alloc: fix potential deadlock on zonelist_update_seq seqlock").
> We have to disable IRQ before making zonelist_update_seq.seqcount odd.
> 

Since we must replace local_irq_save() + write_seqlock() with write_seqlock_irqsave() for
CONFIG_PREEMPT_RT=y case but we must not replace local_irq_save() + write_seqlock() with
write_seqlock_irqsave() for CONFIG_PREEMPT_RT=n case, the proper fix is something like below?

diff --git a/mm/page_alloc.c b/mm/page_alloc.c
index 47421bedc12b..e3e9bd719dcc 100644
--- a/mm/page_alloc.c
+++ b/mm/page_alloc.c
@@ -5798,28 +5798,30 @@ static void per_cpu_pages_init(struct per_cpu_pages *pcp, struct per_cpu_zonesta
 #define BOOT_PAGESET_HIGH	0
 #define BOOT_PAGESET_BATCH	1
 static DEFINE_PER_CPU(struct per_cpu_pages, boot_pageset);
 static DEFINE_PER_CPU(struct per_cpu_zonestat, boot_zonestats);
 
 static void __build_all_zonelists(void *data)
 {
 	int nid;
 	int __maybe_unused cpu;
 	pg_data_t *self = data;
+#ifndef CONFIG_PREEMPT_RT
 	unsigned long flags;
 
 	/*
 	 * Explicitly disable this CPU's interrupts before taking seqlock
 	 * to prevent any IRQ handler from calling into the page allocator
 	 * (e.g. GFP_ATOMIC) that could hit zonelist_iter_begin and livelock.
 	 */
 	local_irq_save(flags);
+#endif
 	/*
 	 * Explicitly disable this CPU's synchronous printk() before taking
 	 * seqlock to prevent any printk() from trying to hold port->lock, for
 	 * tty_insert_flip_string_and_push_buffer() on other CPU might be
 	 * calling kmalloc(GFP_ATOMIC | __GFP_NOWARN) with port->lock held.
 	 */
 	printk_deferred_enter();
 	write_seqlock(&zonelist_update_seq);
 
 #ifdef CONFIG_NUMA
@@ -5852,21 +5854,23 @@ static void __build_all_zonelists(void *data)
 		 * secondary cpus' numa_mem as they come on-line.  During
 		 * node/memory hotplug, we'll fixup all on-line cpus.
 		 */
 		for_each_online_cpu(cpu)
 			set_cpu_numa_mem(cpu, local_memory_node(cpu_to_node(cpu)));
 #endif
 	}
 
 	write_sequnlock(&zonelist_update_seq);
 	printk_deferred_exit();
+#ifndef CONFIG_PREEMPT_RT
 	local_irq_restore(flags);
+#endif
 }
 
 static noinline void __init
 build_all_zonelists_init(void)
 {
 	int cpu;
 
 	__build_all_zonelists(NULL);
 
 	/*


By the way, given

  write_seqlock_irqsave(&zonelist_update_seq, flags);
  <<IRQ>>
    some_timer_function() {
      kmalloc(GFP_ATOMIC);
    }
  <</IRQ>>
  printk_deferred_enter();

scenario in CONFIG_PREEMPT_RT=y case is handled by executing some_timer_function()
on a dedicated kernel thread for IRQs, what guarantees that the kernel thread for
IRQs gives up CPU and the user thread which called write_seqlock() gains CPU until
write_sequnlock() is called? How can the kernel figure out that executing the user
thread needs higher priority than the kernel thread?





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