On 06/15/2015 04:22 PM, John Snow wrote: > The only guidance the AHCI specification gives on memory access is: > "Register accesses shall have a maximum size of 64-bits; 64-bit access > must not cross an 8-byte alignment boundary." > > In practice, a real Q35/ICH9 responds to 1, 2, 4 and 8 byte reads > regardless of alignment. Windows 7 can also be observed making 1 byte > reads to the middle of 32 bit registers. > > Introduce a wrapper to supper unaligned accesses to AHCI.
s/supper/support/
> This wrapper will support aligned 8 byte reads, but will make
> no effort to support unaligned 8 byte reads, which although they
> will work on real hardware, are not guaranteed to work and do
> not appear to be used by either Windows or Linux.
>
> Signed-off-by: John Snow <js...@redhat.com>
> ---
> hw/ide/ahci.c | 21 +++++++++++++++++++--
> 1 file changed, 19 insertions(+), 2 deletions(-)
>
> diff --git a/hw/ide/ahci.c b/hw/ide/ahci.c
> index 9e5d862..55779fb 100644
> --- a/hw/ide/ahci.c
> +++ b/hw/ide/ahci.c
> @@ -331,8 +331,7 @@ static void ahci_port_write(AHCIState *s, int port, int
> offset, uint32_t val)
> }
> }
>
> -static uint64_t ahci_mem_read(void *opaque, hwaddr addr,
> - unsigned size)
> +static uint64_t ahci_mem_read_32(void *opaque, hwaddr addr)
> {
> AHCIState *s = opaque;
> uint32_t val = 0;
> @@ -368,6 +367,24 @@ static uint64_t ahci_mem_read(void *opaque, hwaddr addr,
> }
>
>
> +static uint64_t ahci_mem_read(void *opaque, hwaddr addr, unsigned size)
> +{
> + hwaddr aligned = addr & ~0x3;
> + int ofst = addr - aligned;
> + uint64_t lo = ahci_mem_read_32(opaque, aligned);
> + uint64_t hi;
> +
> + /* if 1/2/4 byte read does not cross 4 byte boundary */
> + if (ofst + size <= 4) {
> + return lo >> (ofst * 8);
> + }
At this point, we could assert(size > 1).
> +
> + /* If the 64bit read is unaligned, we will produce undefined
> + * results. AHCI does not support unaligned 64bit reads. */
> + hi = ahci_mem_read_32(opaque, aligned + 4);
> + return (hi << 32) | lo;
This makes no effort to support an unaligned 2 byte (16bit) or 4 byte
(32bit) read that crosses 4-byte boundary. Is that intentional? I know
it is intentional that you don't care about unaligned 64bit reads;
conversely, while your commit message mentioned Windows doing 1-byte
reads in the middle of 32-bit registers, you didn't mention whether
Windows does unaligned 2- or 4-byte reads. So either the comment should
be broadened, or the code needs further tuning.
--
Eric Blake eblake redhat com +1-919-301-3266
Libvirt virtualization library http://libvirt.org
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