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Message-ID: <a1331c86-dc07-4635-b169-623fcdd11824@paulmck-laptop>
Date: Sat, 11 May 2024 12:37:00 -0700
From: "Paul E. McKenney" <paulmck@...nel.org>
To: John Paul Adrian Glaubitz <glaubitz@...sik.fu-berlin.de>
Cc: Arnd Bergmann <arnd@...db.de>,
Linus Torvalds <torvalds@...ux-foundation.org>,
linux-kernel@...r.kernel.org,
Linux-Arch <linux-arch@...r.kernel.org>,
linux-alpha@...r.kernel.org,
Richard Henderson <richard.henderson@...aro.org>,
Ivan Kokshaysky <ink@...assic.park.msu.ru>,
Matt Turner <mattst88@...il.com>,
Alexander Viro <viro@...iv.linux.org.uk>
Subject: Re: [GIT PULL] alpha: cleanups and build fixes for 6.10
On Sat, May 11, 2024 at 08:49:08PM +0200, John Paul Adrian Glaubitz wrote:
> Hi Paul,
>
> On Fri, 2024-05-10 at 15:28 -0700, Paul E. McKenney wrote:
> > > I'm still against dropping pre-EV56 so quickly without a proper phaseout period.
> > > Why not wait for the next LTS release? AFAIK pre-EV56 support is not broken, is
> > > it?
> >
> > Sadly, yes, it is, and it has been broken in mainline for almost two
> > years.
>
> Could you elaborate what exactly is broken? I'm just trying to understand the reasoning.
First, let's make sure that I completely and correctly understand the
situation.
The pre-EV56 Alphas have no byte store instruction, correct?
If that is in fact correct, what code is generated for a volatile store
to a single byte for those CPUs? For example, for this example?
char c;
...
WRITE_ONCE(c, 3);
The rumor I heard is that the compilers will generate a non-atomic
read-modify-write instruction sequence in this case, first reading the
32-bit word containing that byte into a register, then substituting the
value to be stored into corresponding byte of that register, and finally
doing a 32-bit store from that register.
Is that the case, or am I confused?
Thanx, Paul
PS: Or, if you prefer, this example is equivalent:
volatile char c;
...
c = 3;
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