Thread (58 messages) 58 messages, 12 authors, 2020-07-08

Re: [PATCH 18/18] arm64: lto: Strengthen READ_ONCE() to acquire when CLANG_LTO=y

From: Dave Martin <Dave.Martin@arm.com>
Date: 2020-07-07 10:10:25
Also in: linux-alpha, lkml, virtualization

On Mon, Jul 06, 2020 at 07:35:11PM +0100, Will Deacon wrote:
On Mon, Jul 06, 2020 at 05:08:20PM +0100, Dave Martin wrote:
quoted
On Tue, Jun 30, 2020 at 06:37:34PM +0100, Will Deacon wrote:
quoted
diff --git a/arch/arm64/include/asm/rwonce.h b/arch/arm64/include/asm/rwonce.h
new file mode 100644
index 000000000000..515e360b01a1
--- /dev/null
+++ b/arch/arm64/include/asm/rwonce.h
@@ -0,0 +1,63 @@
+/* SPDX-License-Identifier: GPL-2.0 */
+/*
+ * Copyright (C) 2020 Google LLC.
+ */
+#ifndef __ASM_RWONCE_H
+#define __ASM_RWONCE_H
+
+#ifdef CONFIG_CLANG_LTO
+
+#include <linux/compiler_types.h>
+#include <asm/alternative-macros.h>
+
+#ifndef BUILD_VDSO
+
+#ifdef CONFIG_AS_HAS_LDAPR
+#define __LOAD_RCPC(sfx, regs...)					\
+	ALTERNATIVE(							\
+		"ldar"	#sfx "\t" #regs,				\
^ Should this be here?  It seems that READ_ONCE() will actually read
twice... even if that doesn't actually conflict with the required
semantics of READ_ONCE(), it looks odd.
It's patched at runtime, so it's either LDAR or LDAPR.
Agh ignore me, I somehow failed to sport the ALTERNATIVE().

For my understanding -- my background here is a bit shaky -- the LDAPR
gives us load-to-load order even if there is just a control dependency?

If so (possibly dumb question): why can't we just turn this on
unconditionally?  Is there a significant performance impact?

I'm still confused (or ignorant) though.  If both loads are READ_ONCE()
then switching to LDAPR presumably helps, but otherwise, once the
compiler has reduced the address dependency to a control dependency
can't it then go one step further and reverse the order of the loads?
LDAPR wouldn't rescue us from that.

Or does the "memory" clobber in READ_ONCE() fix that for all important
cases?  I can't see this mattering for local variables (where it
definitely won't work), but I wonder whether static variables might not
count as "memory" in some situations.

Discounting ridiculous things like static register variables, I think
the only way for a static variable not to count as memory would be if
there are no writes to it that are reachable from any translation unit
entry point (possibly after dead code removal).  If so, maybe that's
enough.
quoted
Making a direct link between LTO and the memory model also seems highly
spurious (as discussed in the other subthread) so can we have a comment
explaining the reasoning?
Sure, although like I say, this is more about helping to progress that
conversation.
That's fair enough, but when there is a consensus it would be good to
see it documented in the code _especially_ if we know that the fix won't
address all instances of the problem and in any case works partly by
accident.  That doesn't mean it's not a good practical compromise, but
it could be very confusing to unpick later on.

Cheers
---Dave

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