linux-kernel - Re: [RFC PATCH 2/2] drivers/base: add managed token devres interfaces

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Message-ID: <20140417202221.GU15326@htj.dyndns.org>
Date:	Thu, 17 Apr 2014 16:22:21 -0400
From:	Tejun Heo <tj@...nel.org>
To:	Shuah Khan <shuah.kh@...sung.com>
Cc:	gregkh@...uxfoundation.org, m.chehab@...sung.com,
	rafael.j.wysocki@...el.com, linux@...ck-us.net, toshi.kani@...com,
	linux-kernel@...r.kernel.org, linux-media@...r.kernel.org,
	shuahkhan@...il.com
Subject: Re: [RFC PATCH 2/2] drivers/base: add managed token devres interfaces

On Thu, Apr 17, 2014 at 04:10:34PM -0400, Tejun Heo wrote:
> On Thu, Apr 17, 2014 at 02:01:32PM -0600, Shuah Khan wrote:
> > Operating on the lock should be atomic, which is what devres_update()
> > is doing. It can be simplified as follows by holding devres_lock
> > in devm_token_lock().
> > 
> > spin_lock_irqsave(&dev->devres_lock, flags);
> > if (tkn_ptr->status == TOKEN_DEVRES_FREE)
> > 	tkn_ptr->status = TOKEN_DEVRES_BUSY;
> > spin_unlock_irqrestore(&dev->devres_lock, flags);
> > 
> > Is this in-line with what you have in mind?
> 
> How is that different from tkn_ptr->status = TOKEN_DEVRES_BUSY?

Let me clear it up.  How could the code snippet that you wrote
possibly function as a lock between two threads?  You're doing the
following.


	if (state == busy)
		return -EBUSY;

	spin_lock;
	if (state == free)
		state = busy;
	spin_unlock;

	return SUCCESS!!!11!!1!!;

The above is equivalent to

	if (state == busy)
		return -EBUSY;
	state = busy;
	return SUCCESS!!!11!!1!!;

Now, if you let two threads compete on it, both can return SUCCESS.
Can you see that?

Please consult with somebody who has basic understanding of
concurrency and synchronization.  Please do not implement locking
primitive directly if at all avoidable.  Why can't it use a mutex
embedded in the data area of a devres entry?  And if you for some
reason have to implement it directly, at least add lockdep
annotations.

Anyways, I really think this needs help from somebody who knows
better.

Thanks.

-- 
tejun
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