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Message-ID: <20250226182809.5027e483@gandalf.local.home>
Date: Wed, 26 Feb 2025 18:28:09 -0500
From: Steven Rostedt <rostedt@...dmis.org>
To: Linus Torvalds <torvalds@...ux-foundation.org>
Cc: Martin Uecker <uecker@...raz.at>, Ralf Jung <post@...fj.de>, "Paul E.
 McKenney" <paulmck@...nel.org>, Alice Ryhl <aliceryhl@...gle.com>, Ventura
 Jack <venturajack85@...il.com>, Kent Overstreet
 <kent.overstreet@...ux.dev>, Gary Guo <gary@...yguo.net>,
 airlied@...il.com, boqun.feng@...il.com, david.laight.linux@...il.com,
 ej@...i.de, gregkh@...uxfoundation.org, hch@...radead.org, hpa@...or.com,
 ksummit@...ts.linux.dev, linux-kernel@...r.kernel.org,
 miguel.ojeda.sandonis@...il.com, rust-for-linux@...r.kernel.org
Subject: Re: C aggregate passing (Rust kernel policy)

On Wed, 26 Feb 2025 15:18:48 -0800
Linus Torvalds <torvalds@...ux-foundation.org> wrote:

> On Wed, 26 Feb 2025 at 14:34, Steven Rostedt <rostedt@...dmis.org> wrote:
> >
> > Correct, but if the variable had some other protection, like a lock held
> > when this function was called, it is fine to do and the compiler may
> > optimize it or not and still have the same result.  
> 
> Sure.
> 
> But locking isn't always there. And shouldn't always be there. Lots of
> lockless algorithms exist, and some of them are very simple indeed ("I
> set a flag, you read a flag, you get one or the other value")

Yes, for the case of:

	r = READ_ONCE(global);
	if (r > 1000)
		goto out;
	x = r;

As I've done that in my code without locks, as I just need a consistent
value not necessarily the "current" value.

I was talking for the case the code has (not the compiler creating):

	if (global > 1000)
		goto out;
	x = global;

Because without a lock or some other protection, that's likely a bug.

My point is that the compiler is free to turn that into:

	r = READ_ONCE(global);
	if (r > 1000)
		goto out;
	x = r;

and not change the expected result.

-- Steve


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