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Message-ID: <20150615171848.GP7557@n2100.arm.linux.org.uk>
Date: Mon, 15 Jun 2015 18:18:48 +0100
From: Russell King - ARM Linux <linux@....linux.org.uk>
To: Jon Nettleton <jon.nettleton@...il.com>
Cc: Steffen Trumtrar <s.trumtrar@...gutronix.de>,
Herbert Xu <herbert@...dor.apana.org.au>,
linux-kernel@...r.kernel.org,
Ruchika Gupta <ruchika.gupta@...escale.com>,
linux-crypto@...r.kernel.org, Sascha Hauer <kernel@...gutronix.de>,
linux-arm-kernel@...ts.infradead.org
Subject: Re: [BUG?] crypto: caam: little/big endianness on ARM vs PPC
On Mon, Jun 15, 2015 at 06:33:17PM +0200, Jon Nettleton wrote:
> Funny enough I tackled this problem over the weekend as well. My
> approach was to switch the driver over to use the *_relaxed() io
> functions and then special case the bits missing from the various
> ARCHs. Basically adding setbits32 and clrbits32 for !PPC
> architectures and letting PPC and ARM share a writeq/readq set of
> functions. I left the existing LITTLE_ENDIAN special case until I
> could verify if it was needed, or had been tested.
I'll follow up here with what I've mentioned elsewhere, and some further
thoughts.
I think this shows the dangers of using struct { } to define register
offsets. Let's start here:
/*
* caam_job_ring - direct job ring setup
* 1-4 possible per instantiation, base + 1000/2000/3000/4000
* Padded out to 0x1000
*/
struct caam_job_ring {
/* Input ring */
u64 inpring_base; /* IRBAx - Input desc ring baseaddr */
u32 rsvd1;
Apparently, this is a CPU-endian 64-bit value (it's not defined using
le64 or be64 which would "fix" it's endian.)
The second question, which comes up in light of the breakage that's
being reported is: is this really a 64-bit register, or is it a pair
of 32-bit registers side-by-side?
The documentation I'm looking at doesn't document the register at
base + 0x1000, but documents the one at base + 0x1004, and the one
at 0x1004 is given the name "IRBAR0_LS", which presumably stands
for "input ring base address register 0, least significant".
As the code originally stood for PPC, IRBAR0_LS is also at 0x1004,
but appears to be big endian.
On ARM, IRBAR0_LS appears at the same address, but is little endian.
This is *not* a 64-bit register at all, but is a pair of 32-bit
registers side by side. Moreover, readq() should not be used - no
amount of arch mangling could ever produce a sane readq() which
coped with this.
So, the CAAM code is buggy in this regard: using readq() here when
endian-portability is required is wrong. It's got to be two 32-bit
reads or two 32-bit writes in the appropriate endian.
Also, note that there's a big difference between __raw_readl() and
readl_relaxed(). readl_relaxed() is always little-endian. __raw_readl()
is god-knows-what-the-archtecture-decides endian. Switching PPC
drivers from __raw_readl() to readl_relaxed() is really not a good
idea unless someone from the PPC camp reviews and tests the code.
So, what I'd suggest is just fixing rd_reg64() and wr_reg64() to do
the right thing: keeping the two 32-bit words in the same order
irrespective of the endian-ness, and staying with the __raw_*
accessors until PPC people can look at this.
--
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according to speedtest.net.
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