On Thu, Oct 18, 2012 at 06:48:16AM +0100, Geert Uytterhoeven wrote:
> On Thu, Oct 18, 2012 at 2:04 AM, Benjamin Herrenschmidt
> <b...@kernel.crashing.org> wrote:
> > The sort story is that endianness is not a property of the IO port but
> > of the information that transit through it. If you're just going to copy
> > it into memory, you want to preserve it's format and so do not byteswap.
> >
> > The byteswap we do on standard accessors is a "helper" because we assume
> > that underneath those IO ports are registers that are Little Endian. But
> > when using one as a window to a byte stream, we must not arbitrarily
> > swap the byte stream. We copy it as-is to memory, and then one can work
> > at interpreting the various fields that might or might not be present in
> > that stream with the appropriate accessors for memory accesses.
>
> So assume you have the bytestream "Hello, world!\n" in memory on the
> PCI device.I.e.
>
> 00000000 48 65 6c 6c 6f 2c 20 77 6f 72 6c 64 21 0a |Hello, world!.|
>
> You want to copy it to system RAM using readsl(), which does:
>
> u32 *buf = buffer;
> do {
> u32 x = __raw_readl(addr + PCI_IOBASE);
> *buf++ = x;
> } while (--count);
>
> On little endian, the first __raw_readl() should return "0x6c6c6548", so
> it is stored correctly by "*buf = x ".
> On big endian, the first __raw_readl() should return "0x48656c6c" instead,
> else it's stored incorrectly by "*buf = x ".
So far so good...
> But the PCI bus is little endian, so I expect __raw_readl() would return
> "0x6c6c6548", and thus needs swapping?
I think this would only happen if your busses are wired swapped, in which
case you'll have to handle this in your arch code because reading from a
device and then writing to memory will end up with the data in the wrong
order (the data stream won't be affected by passing through the CPU).
Will
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