Re: [PATCH] mlx4_en: fix transmit of packages when blue frame is enabled

[Benjamin Herrenschmidt <benh@kernel.crashing.org>]

On Sun, 2011-10-09 at 09:35 +0200, Eli Cohen wrote:
> On Sun, Oct 09, 2011 at 09:25:18AM +0200, Benjamin Herrenschmidt wrote:
> > On Thu, 2011-10-06 at 15:57 +0200, Eli Cohen wrote:
> > > On Wed, Oct 05, 2011 at 10:15:02AM +0200, Eli Cohen wrote:
> > > 
> > > How about this patch - can you give it a try?
> > > 
> > > 
> > > >From dee60547aa9e35a02835451d9e694cd80dd3072f Mon Sep 17 00:00:00 2001
> > > From: Eli Cohen <eli@mellanox.co.il>
> > > Date: Thu, 6 Oct 2011 15:50:02 +0200
> > > Subject: [PATCH] mlx4_en: Fix blue flame on powerpc
> > > 
> > > The source buffer used for copying into the blue flame register is already in
> > > big endian. However, when copying to device on powerpc, the endianess is
> > > swapped so the data reaches th device in little endian which is wrong. On x86
> > > based platform no swapping occurs so it reaches the device with the correct
> > > endianess. Fix this by calling le32_to_cpu() on the buffer. On LE systems there
> > > is no change; on BE there will be a swap.
> > 
> > That looks wrong.
> Not sure I understand: are you saying that on ppc, when you call
> __iowrite64_copy, it will not reach the device swapped?

Well, first, what do you mean by "swapped" ? :-) But no, it won't for
all intend and purpose, this is a copy routine, copy routines never
swap, neither do fifo accesses for example.

> The point is that we must always have the buffer ready in big endian
> in memory. In the case of blue flame, we must also copy it to the
> device registers in pci memory space. So if we use the buffer we
> already prepared, we must have another swap. I can think of a nicer
> way to implement this functionality but the question is do you think
> my observation above is wrong and why.

No. If it's in memory BE then the copy routine will keep it BE. A copy
routine doesn't swap and doesn't affect endianness.

Additionally, a swapping phase like you proposed doing 32-bit swaps
means that you know for sure that the buffer is made of 32-bit
quantities, is that the case ? Even if you had needed that swap, if your
buffer had contained 16-bit or 64-bit quantities, you're toast.

What is this buffer anyway ? A descriptor or a network packet ?

If it's a packet, then it's data, endianness has no meaning (or rather
it has for individual fields of the packets but they are already in the
right format and a 32-bit swap will never be right).
 
It's almost never right to perform swapping when copying data (or
reading/writing a FIFO).

Cheers,
Ben.