From mboxrd@z Thu Jan  1 00:00:00 1970
From: erik quanstrom <quanstro@quanstro.net>
Date: Thu,  2 Jan 2014 17:57:54 -0500
To: 9fans@9fans.net
Message-ID: <a479f284726ba7c37747d63551fc4c4d@mikro>
In-Reply-To: <88f3a9a7b4b349bbc153cae9e1f0dfde@felloff.net>
References: <88f3a9a7b4b349bbc153cae9e1f0dfde@felloff.net>
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Subject: Re: [9fans] devproc noteid changing for none
Topicbox-Message-UUID: ab793ef2-ead8-11e9-9d60-3106f5b1d025

On Thu Jan  2 17:54:50 EST 2014, cinap_lenrek@felloff.net wrote:
> no. the debug qlock is so the process wont exit. once you hold it,
> the process might be running, might be in the process of exiting
> might have been already exited or might have been reused.
>
> the key here is that while you hold it. it wont make it to the
> process freelist when not already there. it wont change
> from "exiting -> exited". but it wont prevent it from being
> reused when it was exited already... but...
>
> pexit() zeros p->alarm. and p->alarm can not be set to non-zero
> value again while we hold the alarms qlock in the kproc!
>
> so this actually works.
>

the code says you can set alarm = 0 without holding the alarms lock.

ulong
procalarm(ulong time)
{
	Proc **l, *f;
	ulong when, old;

	if(up->alarm)
		old = tk2ms(up->alarm - sys->ticks);
	else
		old = 0;
	if(time == 0) {
>>		up->alarm = 0;
>>		return old;

- erik