On Fri, Aug 25, 2006 at 05:57:24PM -0400, Lennart Sorensen wrote:
> It would be 1/4 of the time given the code is this:
>
> memcpy_toio(&ch->ch_neo_uart->txrxburst, ch->ch_wqueue + tail, s);
>
> tail is the offset in the write queue in the jsm driver, which can be
> any offset whatsoever. So sometimes it is 32bit aligned, but often it
> isn't.
>
> The txrxburst of course is 32bit alligned. Receiving works fine when
> copying from the alligned buffer in hardware to wherever in the receiver
> queue.
>
> Of course given the __memcpy assembly seems to work fine unalligned on a
> pentium4, and probably most othe systems, what could make it not work
> correctly on a geode SC1200?
>
> This is the chunk of assembly in use:
>
> static __always_inline void * __memcpy(void * to, const void * from, size_t n)
> {
> int d0, d1, d2;
> __asm__ __volatile__(
> "rep ; movsl\n\t"
> "movl %4,%%ecx\n\t"
> "andl $3,%%ecx\n\t"
> #if 1 /* want to pay 2 byte penalty for a chance to skip microcoded rep? */
> "jz 1f\n\t"
> #endif
> "rep ; movsb\n\t"
> "1:"
> : "=&c" (d0), "=&D" (d1), "=&S" (d2)
> : "0" (n/4), "g" (n), "1" ((long) to), "2" ((long) from)
> : "memory");
> return (to);
> }
>
> I am affraid I don't know the rep instruction on x86, so it really
> doesn't make sense to me.
>
> I suppose if nothing else works, I can do one byte at a time until the
> tail is 32bit alligned, and then do the rest of the transfer as a block
> and see if that makes it work, or whether it is broken no matter what
> the allignement of the buffer is.
I tried to do a check on the alignment, and only do the whole transfer
as a block with memcpy_toio if it was 32bit aligned. It breaks the same
way no matter how it is aligned. It really seems that somehow using
this __memcpy assembly block on the geode sc1200 doesn't work correctly.
Doing the call using a loop with d2=1 works fine, but anything bigger
seems broken.
--
Len Sorensen
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