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Message-ID: <alpine.LNX.2.00.1012011536520.8521@localhost.localdomain>
Date: Wed, 1 Dec 2010 16:02:14 +0100 (CET)
From: Hans Ulli Kroll <ulli.kroll@...glemail.com>
To: Paulius Zaleckas <paulius.zaleckas@...il.com>
cc: Hans Ulli Kroll <ulli.kroll@...glemail.com>,
Arnd Bergmann <arnd@...db.de>,
linux-arm-kernel@...ts.infradead.org,
Russell King <linux@....linux.org.uk>,
linux-kernel@...r.kernel.org
Subject: Re: [PATCH] ARM: Gemini: Add support for PCI BUS
On Wed, 1 Dec 2010, Paulius Zaleckas wrote:
> On Wed, Dec 1, 2010 at 1:52 PM, Hans Ulli Kroll
> <ulli.kroll@...glemail.com> wrote:
> >
> >
> > On Tue, 30 Nov 2010, Paulius Zaleckas wrote:
> >
> >> On Tue, Nov 30, 2010 at 10:15 AM, Hans Ulli Kroll
> >> <ulli.kroll@...glemail.com> wrote:
> >> >
> >> >
> >> > On Mon, 29 Nov 2010, Paulius Zaleckas wrote:
> >> >
> >> >> On 11/29/2010 10:02 PM, Arnd Bergmann wrote:
> >> >> > On Monday 29 November 2010 19:52:55 Paulius Zaleckas wrote:
> >> >> > > On 11/29/2010 06:45 PM, Arnd Bergmann wrote:
> >> >> > > > There are many differences between readl and __raw_readl, including
> >> >> > > >
> >> >> > > > * __raw_readl does not have barriers and does not serialize with
> >> >> > > > spinlocks, so it breaks on out-of-order CPUs.
> >> >> > > > * __raw_readl does not have a specific endianess, while readl is
> >> >> > > > fixed little-endian, just as the hardware is in this case.
> >> >> > > > The endian-conversion is a NOP on little-endian ARM, but required
> >> >> > > > if you actually run on a big-endian ARM (you don't).
> >> >> > > > * __raw_readl may not be atomic, gcc is free to split the access
> >> >> > > > into byte wise reads (it normally does not, unless you mark
> >> >> > > > the pointer __attribute__((packed))).
> >> >> > > >
> >> >> > > > In essence, it is almost never a good idea to use __raw_readl, and
> >> >> > > > the double underscores should tell you so.
> >> >> > >
> >> >> > > You are wrong:
> >> >> > >
> >> >> > > Since CONFIG_ARM_DMA_MEM_BUFFERABLE is NOT defined for FA526 core,
> >> >> > > no barriers are in use when using readl. It just translates into
> >> >> > > le32_to_cpu(__raw_readl(x)). Now this CPU has physical pin for endianess
> >> >> > > configuration and if you will chose big-endian you will fail to read
> >> >> > > internal registers, because they ALSO change endianess and le32_to_cpu()
> >> >> > > will screw it. However it is different when accessing registers through
> >> >> > > PCI bus, then you need to use readl().
> >> >> >
> >> >> > Ok, I only checked that the platform does not support big-endian Linux
> >> >> > kernel, not if the HW designers screwed up their registers, sorry about
> >> >> > that.
> >> >> >
> >> >> > The other points are of course still valid: If the code ever gets
> >> >> > used on an out of order CPU, it is broken. More importantly, if someone
> >> >> > looks at the code as an example for writing another PCI support code,
> >> >> > it may end up getting copied to some place where it ends up causing
> >> >> > trouble.
> >> >> >
> >> >> > The typical way to deal with mixed-endian hardware reliably is to have
> >> >> > a header file containing code like
> >> >> >
> >> >> > #ifdef CONFIG_GEMINI_BIG_ENDIAN_IO
> >> >> > #define gemini_readl(x) __swab32(readl(x))
> >> >> > #define ...
> >> >> > #else
> >> >> > #define gemini_readl(x) readl(x))
> >> >> > #endif
> >> >> >
> >> >> > This also takes care of the (not as unlikely as you'd hope) case that
> >> >> > the next person reusing the PCI hardware wires its endianess different
> >> >> > from the CPU endianess.
> >> >>
> >> >> Actually I am not very sure how CPU works in big endian mode :)
> >> >> I have never tried it and I think only some guys who made it did that.
> >> >> So readl will work for 99.99% of cases. In datasheet they say that:
> >> >> "All registers in Gemini use Little Endian and must be accessed by aligned
> >> >> 32-bit word operations. The bus connection interface logic provides an Endian
> >> >> Conversion function."
> >> >> For me it looks like it can mean whatever you want :)
> >> >>
> >> >
> >> > I think the endianes pin switched only the cpu, not the hardware
> >> > registers.
> >>
> >> Yes, but original driver used readl/writel and it does le32_to_cpu,
> >> so that structure definition is just reversing it.
> >> If you will use __raw_readl/__raw_writel than there will be no need
> >> for this redefinition.
> >>
> >> > Here is some sample code from the ethernet devive on Gemini
> >> > typedef union
> >> > {
> >> > unsigned int bits32;
> >> > struct bit
> >> > {
> >> > #if (BIG_ENDIAN==1)
> >> > unsigned int reserved : 15; // bit 31:17
> >> > unsigned int v_bit_mode : 1; // bit 16
> >> > unsigned int device_id : 12; // bit 15:4
> >> > unsigned int revision_id : 4; // bit 3:0
> >> > #else
> >> > unsigned int revision_id : 4; // bit 3:0
> >> > unsigned int device_id : 12; // bit 15:4
> >> > unsigned int v_bit_mode : 1; // bit 16
> >> > unsigned int reserved : 15; // bit 31:17
> >> > #endif
> >>
> >> The other thing is that this endianess redefinition is very starnge since
> >> it should swap bytes and not bits inside this struct. So I assume that
> >> big endian was never tested on this driver and it will not work.
> >> But ofcouse I can be wrong here :)
> >>
> >
> > At this momment my brain restarts in very slow motion mode ;-)
> > This can't work. The definition Storlinksemi uses for swapping bits and
> > bytes are totaly wrong.
> > They never _even_ testet this, or understand little endian or big endian.
> > Take this simple sample
> >
> > typedef union {
> > unsigned int bits32;
> > struct bit {
> > #if (BIG_ENDIAN==0)
> > unsigned int a : 1;
> > unsigned int b : 31;
> > #else
> > unsigned int b : 31;
> > unsigned int a : 1;
> > #endif
> > };
> > } TEST;
> >
> > They swaped the bits inside one byte
>
> Ha! Wait a minute. Looks like we are both wrong...
> Read the beginning of: http://thread.gmane.org/gmane.linux.kernel/1037608
>
> So it means we should use readl/writel and get rid of these non-portable
> bit-fields...
>
get rid of _endianes_.
Endianes is ordering of bytes (octets) in address space and _not_ bits in
registers nor bytes.
With the above example :
a = 1;
b = 0;
you will get on little endian cpu this
0x01, 0x00, 0x00, 0x00 (0x00000001UL) in memory
on big endian
0x00, 0x00, 0x00, 0x01 (0x01000000UL)
_note_ the bits are not swapped, _only_ bytes.
but if we use this _strange_ register layout with BIG_ENDIAN is set
0x00, 0x00, 0x00, 0x80 (0x80000000UL)
I've done this _on_ paper to understand this.
> >> > } bits;
> >> > } TOE_VERSION_T;
> >> >
> >> >
> >> >
> >>
>
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