linux-kernel - Re: [PATCH 2/2] cgroup-v2/freezer: Print information about unfreezable process

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Message-ID: <80d36215-de7e-448c-85ea-9e848496c210@virtuozzo.com>
Date: Mon, 29 Dec 2025 13:32:11 +0800
From: Pavel Tikhomirov <ptikhomirov@...tuozzo.com>
To: Tejun Heo <tj@...nel.org>
Cc: Johannes Weiner <hannes@...xchg.org>, Michal Koutný
 <mkoutny@...e.com>, cgroups@...r.kernel.org, linux-kernel@...r.kernel.org
Subject: Re: [PATCH 2/2] cgroup-v2/freezer: Print information about
 unfreezable process



On 12/28/25 06:51, Tejun Heo wrote:
> Hello,
> 
> On Tue, Dec 23, 2025 at 06:20:09PM +0800, Pavel Tikhomirov wrote:
>> +static void warn_freeze_timeout(struct cgroup *cgrp, int timeout)
>> +{
>> +	char *buf __free(kfree) = NULL;
>> +	struct cgroup_subsys_state *css;
>> +
>> +	guard(rcu)();
>> +	css_for_each_descendant_post(css, &cgrp->self) {
>> +		struct task_struct *task;
>> +		struct css_task_iter it;
>> +
>> +		css_task_iter_start(css, 0, &it);
>> +		while ((task = css_task_iter_next(&it))) {
>> +			if (task->flags & PF_KTHREAD)
>> +				continue;
>> +			if (task->frozen)
>> +				continue;
>> +
>> +			warn_freeze_timeout_task(cgrp, timeout, task);
>> +			css_task_iter_end(&it);
>> +			return;
>> +		}
>> +		css_task_iter_end(&it);
>> +	}
>> +
>> +	buf = kmalloc(PATH_MAX, GFP_KERNEL);
>> +	if (!buf)
>> +		return;
>> +
>> +	if (cgroup_path(cgrp, buf, PATH_MAX) < 0)
>> +		return;
>> +
>> +	pr_warn("Freeze of %s took %ld sec, but no unfreezable process detected.\n",
>> +		buf, timeout / USEC_PER_SEC);
>> +}
> 
> This is only suitable for debugging, and, for that, this can be done from
> userspace by walking the tasks and check /proc/PID/wchan. Should be
> do_freezer_trap for everything frozen. If something is not, read and dump
> its /proc/PID/stack. Wouldn't that work?

Yes, I think that will do, Thanks.

Though the trace printing in /proc/PID/stack is a bit less informative than show_stack(), e.g. for my test module (https://github.com/Snorch/proc-hang-module) the stack in /proc/PID/stack will be just empty. (I guess it has to do with the fact that you need less privileges to read /proc/PID/stack than dmesg, so you can do a more informative thing in dmesg).

> 
> Thanks.
> 

-- 
Best regards, Pavel Tikhomirov
Senior Software Developer, Virtuozzo.