Re: why it not write those 6bits to entrylo0/1 register?

[figo zhang <figo1802@gmail.com>]

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>
> No, the low 6 bits contain other information maintained by the kernel.
> Shifting right by 6 bits is used to drop these software bits.  The
> hardware bits are stored in bits 6 and up in a pte so the shift operation
> is going to move them into the right place.
>

yes, i know why shout shift this 6 bits, see this :

entrylo[01]:
  3130 29                         6 5 3 2 1 0
  -------------------------------------------
  | | PFN                         | C |D|V|G|
  -------------------------------------------

linux pte:
  31                     12 111098 7 6 5 3 2 1 0
  -------------------------------------------
  | PFN                | C |D|V|G|B|M|A|W|R|P|
  -------------------------------------------

so , the linux PTE has the least significant 6 bits is mantain by linux PTE,
the hardware
PTE entrylo[0~1] have no such bits, so it need to shift .

ralf, is some description on the kernel code? if it has, it would be easy
understand .

>
> > D:
> > V:
> > G:
> >
> > and how the kernel write the this 6 bit to entrylo0/1 register?
>
> A TLB write instruction about 5 lines further down in the code.
>
>  Ralf
>

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