Justin Fletcher wrote:
>
> I may be confused on this, but it still doesn't seem right to me.
>
> You have...
>
> - ap = (desc >> (4 + ((address >> 13) & 6))) & 3;
> + ap = (desc >> (4 + ((address >> 11) & 6))) & 3; /* SRO */
>
> For 4K pages, the L2 table is ...
> b0-1 = 2
> b2 = B
> b3 = C
> b4-5 = AP0
> b6-7 = AP1
> b8-9 = AP2
> b10-11=AP3
> b12-31=physical address
> (from ARMARM 'D', 3.3.7)
>
> The use of AP0-AP3 is dependant on bits 10 and 11. So, the code should
> be more like...
>
> ap = (desc >> (4 + ((address >> 10) & 3) )) & 3;
>
> That is, (address>>10) & 3 => bits 10 and 11
> add on 4 as the offset to the AP fields in the descriptor
> shift down and & 3 to leave just those two bits.
>
Well, we need to take b10-11 and use them to index either 4-5, 6-7, 8-9
or 10-11.
(address >> 10) & 3 gives us 0, 1, 2 or 3, shift that left one to double
it (because each AP field is two bits). Adding 4 gives 4, 6, 8, 10. So
I believe the correct solution is:
ap = (desc >> (4 + ((address >> 9) & 6))) & 3;
I thought if was just 2 bits different from the large page descriptor,
but the difference, the SBZ field, is 4 bits. Comparing to the large
page descriptor:
ap = (desc >> (4 + ((address >> 13) & 6))) & 3;
-Scott
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