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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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