Hi Suzuki,
On 06/29/2018 01:15 PM, Suzuki K Poulose wrote:
On arm64 VTTBR_EL2:BADDR holds the base address for the stage2Isn't it a pretty old reference? Could you refer to C.a?
translation table. The Arm ARM mandates that the bits BADDR[x-1:0]
should be 0, where 'x' is defined for a given IPA Size and the
number of levels for a translation granule size. It is defined
using some magical constants. This patch is a reverse engineered
implementation to calculate the 'x' at runtime for a given ipa and
number of page table levels. See patch for more details.
Cc: Marc Zyngier <marc.zyngier@xxxxxxx>
Cc: Christoffer Dall <cdall@xxxxxxxxxx>
Signed-off-by: Suzuki K Poulose <suzuki.poulose@xxxxxxx>
---
Changes since V2:
- Part 1 of spilt from VTCR & VTTBR dynamic configuration
---
arch/arm64/include/asm/kvm_arm.h | 60 +++++++++++++++++++++++++++++++++++++---
arch/arm64/include/asm/kvm_mmu.h | 25 ++++++++++++++++-
2 files changed, 80 insertions(+), 5 deletions(-)
diff --git a/arch/arm64/include/asm/kvm_arm.h b/arch/arm64/include/asm/kvm_arm.h
index 3dffd38..c557f45 100644
--- a/arch/arm64/include/asm/kvm_arm.h
+++ b/arch/arm64/include/asm/kvm_arm.h
@@ -140,8 +140,6 @@
* Note that when using 4K pages, we concatenate two first level page tables
* together. With 16K pages, we concatenate 16 first level page tables.
*
- * The magic numbers used for VTTBR_X in this patch can be found in Tables
- * D4-23 and D4-25 in ARM DDI 0487A.b.
+ *I understand entry level = Lookup level in the table.
+ * The algorithm defines the expectations on the BaseAddress (for the page
+ * table) bits resolved at each level based on the page size, entry level
+ * and T0SZ. The variable "x" in the algorithm also affects the VTTBR:BADDR
+ * for stage2 page table.
+ *
+ * The value of "x" is calculated as :
+ * x = Magic_N - T0SZ
+ *
+ * where Magic_N is an integer depending on the page size and the entry
+ * level of the page table as below:
+ *
+ * --------------------------------------------
+ * | Entry level | 4K 16K 64K |
+ * --------------------------------------------
+ * | Level: 0 (4 levels) | 28 | - | - |
+ * --------------------------------------------
+ * | Level: 1 (3 levels) | 37 | 31 | 25 |
+ * --------------------------------------------
+ * | Level: 2 (2 levels) | 46 | 42 | 38 |
+ * --------------------------------------------
+ * | Level: 3 (1 level) | - | 53 | 51 |
+ * --------------------------------------------
But you may want to compute x for BaseAddress matching lookup level 2
with number of levels = 4.
So shouldn't you s/Number of levels/4 - entry_level?
for BADDR we want the BaseAddr of the initial lookup level so
effectively the entry level we are interested in is 4 - number of levels
and we don't care or d) condition. At least this is my understanding ;-)
If correct you may slightly reword the explanation?
+ *
+ * We have a magic formula for the Magic_N below.
+ *
+ * Magic_N(PAGE_SIZE, Entry_Level) = 64 - ((PAGE_SHIFT - 3) * Number of levels)
+ *
+ * where number of levels = (4 - Entry_Level).
+/*Link to the spec?
+ * Get the magic number 'x' for VTTBR:BADDR of this KVM instance.
+ * With v8.2 LVA extensions, 'x' should be a minimum of 6 with
+ * 52bit IPS.