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Date:   Fri, 22 Mar 2019 12:16:42 +0100
From:   Oleg Nesterov <>
To:     Matthew Wilcox <>
Cc:     Waiman Long <>,
        Andrew Morton <>,
        Christoph Lameter <>,
        Pekka Enberg <>,
        David Rientjes <>,
        Joonsoo Kim <>,,,, Paul Moore <>,
        Stephen Smalley <>,
        Eric Paris <>,
        "Peter Zijlstra (Intel)" <>
Subject: Re: [PATCH 2/4] signal: Make flush_sigqueue() use free_q to release

On 03/21, Matthew Wilcox wrote:
> On Thu, Mar 21, 2019 at 05:45:10PM -0400, Waiman Long wrote:
> > To avoid this dire condition and reduce lock hold time of tasklist_lock,
> > flush_sigqueue() is modified to pass in a freeing queue pointer so that
> > the actual freeing of memory objects can be deferred until after the
> > tasklist_lock is released and irq re-enabled.
> I think this is a really bad solution.  It looks kind of generic,
> but isn't.  It's terribly inefficient, and all it's really doing is
> deferring the debugging code until we've re-enabled interrupts.


> We'd be much better off just having a list_head in the caller
> and list_splice() the queue->list onto that caller.  Then call
> __sigqueue_free() for each signal on the queue.

This won't work, note the comment which explains the race with sigqueue_free().

Let me think about it... at least we can do something like

	close_the_race_with_sigqueue_free(struct sigpending *queue)
		struct sigqueue *q, *t;

		list_for_each_entry_safe(q, t, ...) {
			if (q->flags & SIGQUEUE_PREALLOC)

called with ->siglock held, tasklist_lock is not needed.

After that flush_sigqueue() can be called lockless in release_task() release_task.

I'll try to make the patch tomorrow.


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