Re: [mm/contpte v3 1/1] mm/contpte: Optimize loop to reduce redundant operations

[Barry Song]

On Tue, Apr 22, 2025 at 9:34 PM Xavier <xavier_qy@xxxxxxx> wrote:
>
>
> Hi all,
>
>
> At 2025-04-16 20:54:47, "Ryan Roberts" <ryan.roberts@xxxxxxx> wrote:
> >On 15/04/2025 09:22, Xavier wrote:
> >> This commit optimizes the contpte_ptep_get function by adding early
> >>  termination logic. It checks if the dirty and young bits of orig_pte
> >>  are already set and skips redundant bit-setting operations during
> >>  the loop. This reduces unnecessary iterations and improves performance.
> >>
> >> Signed-off-by: Xavier <xavier_qy@xxxxxxx>
> >> ---
> >>  arch/arm64/mm/contpte.c | 20 ++++++++++++++++++--
> >>  1 file changed, 18 insertions(+), 2 deletions(-)
> >>
> >> diff --git a/arch/arm64/mm/contpte.c b/arch/arm64/mm/contpte.c
> >> index bcac4f55f9c1..0acfee604947 100644
> >> --- a/arch/arm64/mm/contpte.c
> >> +++ b/arch/arm64/mm/contpte.c
> >> @@ -152,6 +152,16 @@ void __contpte_try_unfold(struct mm_struct *mm, unsigned long addr,
> >>  }
> >>  EXPORT_SYMBOL_GPL(__contpte_try_unfold);
> >>
> >> +/* Note: in order to improve efficiency, using this macro will modify the
> >> + * passed-in parameters.*/
> >> +#define CHECK_CONTPTE_FLAG(start, ptep, orig_pte, flag) \
> >> +    for (; (start) < CONT_PTES; (start)++, (ptep)++) { \
> >> +            if (pte_##flag(__ptep_get(ptep))) { \
> >> +                            orig_pte = pte_mk##flag(orig_pte); \
> >> +                            break; \
> >> +            } \
> >> +    }
> >
> >I'm really not a fan of this macro, it just obfuscates what is going on. I'd
> >personally prefer to see the 2 extra loops open coded below.
> >
> >Or even better, could you provide results comparing this 3 loop version to the
> >simpler approach I suggested previously? If the performance is similar (which I
> >expect it will be, especially given Barry's point that your test always ensures
> >the first PTE is both young and dirty) then I'd prefer to go with the simpler code.
> >
>
> Based on the discussions in the previous email, two modifications were adopted
> and tested, and the results are as follows:
>
> Modification 1
>
> pte_t contpte_ptep_get(pte_t *ptep, pte_t orig_pte)
> {
>         pte_t pte;
>         int i;
>
>         ptep = contpte_align_down(ptep);
>
>         for (i = 0; i < CONT_PTES; i++, ptep++) {
>                 pte = __ptep_get(ptep);
>
>                 if (pte_dirty(pte)) {
>                         orig_pte = pte_mkdirty(orig_pte);
>                         if (pte_young(orig_pte))
>                                 break;
>                 }
>
>                 if (pte_young(pte)) {
>                         orig_pte = pte_mkyoung(orig_pte);
>                         if (pte_dirty(orig_pte))
>                                 break;
>                 }
>         }
>
>         return orig_pte;
> }
>
> Modification 2
>
> pte_t contpte_ptep_get(pte_t *ptep, pte_t orig_pte)
> {
>         pte_t pte;
>         int i;
>
>         ptep = contpte_align_down(ptep);
>
>         for (i = 0; i < CONT_PTES; i++, ptep++) {
>                 pte = __ptep_get(ptep);
>
>                 if (pte_dirty(pte)) {
>                         orig_pte = pte_mkdirty(orig_pte);
>                         for (; i < CONT_PTES; i++, ptep++) {
>                                 pte = __ptep_get(ptep);
>                                 if (pte_young(pte)) {
>                                         orig_pte = pte_mkyoung(orig_pte);
>                                         break;
>                                 }
>                         }
>                         break;
>                 }
>
>                 if (pte_young(pte)) {
>                         orig_pte = pte_mkyoung(orig_pte);
>                         i++;
>                         ptep++;
>                         for (; i < CONT_PTES; i++, ptep++) {
>                                 pte = __ptep_get(ptep);
>                                 if (pte_dirty(pte)) {
>                                         orig_pte = pte_mkdirty(orig_pte);
>                                         break;
>                                 }
>                         }
>                         break;
>                 }
>         }
>
>         return orig_pte;
> }
>
> Test Code:
>
> #define PAGE_SIZE 4096
> #define CONT_PTES 16
> #define TEST_SIZE (4096* CONT_PTES * PAGE_SIZE)
> #define YOUNG_BIT 8
> void rwdata(char *buf)
> {
>         for (size_t i = 0; i < TEST_SIZE; i += PAGE_SIZE) {
>                 buf[i] = 'a';
>                 volatile char c = buf[i];
>         }
> }
> void clear_young_dirty(char *buf)
> {
>         if (madvise(buf, TEST_SIZE, MADV_FREE) == -1) {
>                 perror("madvise free failed");
>                 free(buf);
>                 exit(EXIT_FAILURE);
>         }
>         if (madvise(buf, TEST_SIZE, MADV_COLD) == -1) {
>                 perror("madvise free failed");
>                 free(buf);
>                 exit(EXIT_FAILURE);
>         }
> }
> void set_one_young(char *buf)
> {
>         for (size_t i = 0; i < TEST_SIZE; i += CONT_PTES * PAGE_SIZE) {
>                 volatile char c = buf[i + YOUNG_BIT * PAGE_SIZE];
>         }
> }
>
> void test_contpte_perf() {
>         char *buf;
>         int ret = posix_memalign((void **)&buf, CONT_PTES * PAGE_SIZE, TEST_SIZE);
>         if ((ret != 0) || ((unsigned long)buf % CONT_PTES * PAGE_SIZE)) {
>                 perror("posix_memalign failed");
>                 exit(EXIT_FAILURE);
>         }
>
>         rwdata(buf);
> #if TEST_CASE2 || TEST_CASE3
>         clear_young_dirty(buf);
> #endif
> #if TEST_CASE2
>         set_one_young(buf);
> #endif
>
>         for (int j = 0; j < 500; j++) {
>                 mlock(buf, TEST_SIZE);
>
>                 munlock(buf, TEST_SIZE);
>         }
>         free(buf);
> }
> ---
>
> Descriptions of three test scenarios
>
> Scenario 1
> The data of all 16 PTEs are both dirty and young.
> #define TEST_CASE2 0
> #define TEST_CASE3 0
>
> Scenario 2
> Among the 16 PTEs, only the 8th one is young, and there are no dirty ones.
> #define TEST_CASE2 1
> #define TEST_CASE3 0
>
> Scenario 3
> Among the 16 PTEs, there are neither young nor dirty ones.
> #define TEST_CASE2 0
> #define TEST_CASE3 1
>
>
> Test results
>
> |Scenario 1         |       Original|  Modification 1|  Modification 2|
> |-------------------|---------------|----------------|----------------|
> |instructions       |    37912436160|     18303833386|     18731580031|
> |test time          |         4.2797|          2.2687|          2.2949|
> |overhead of        |               |                |                |
> |contpte_ptep_get() |         21.31%|           4.72%|           4.80%|
>
> |Scenario 2         |       Original|  Modification 1|  Modification 2|
> |-------------------|---------------|----------------|----------------|
> |instructions       |    36701270862|     38729716276|     36115790086|
> |test time          |         3.2335|          3.5732|          3.0874|
> |Overhead of        |               |                |                |
> |contpte_ptep_get() |         32.26%|          41.35%|          33.57%|
>
> |Scenario 3         |       Original|  Modification 1|  Modification 2|
> |-------------------|---------------|----------------|----------------|
> |instructions       |    36706279735|     38305241759|     36750881878|
> |test time          |         3.2008|          3.5389|          3.1249|
> |Overhead of        |               |                |                |
> |contpte_ptep_get() |         31.94%|          41.30%|          34.59%|
>
>
> For Scenario 1, Modification 1 can achieve an instruction count benefit of
> 51.72% and a time benefit of 46.99%. Modification 2 can achieve an instruction
> benefit of 50.59% and a time benefit of 46.38%.
>
> For Scenarios 2, Modification 2 can achieve an instruction count benefit of
> 1.6% and a time benefit of 4.5%. while Modification 1 significantly increases
> the instructions and time due to additional conditional checks.
>
> For Scenario 3, since all the PTEs have neither the young nor the dirty flag,
> the branches taken by Modification 1 and Modification 2 should be the same as
> those of the original code. In fact, the test results of Modification 2 seem
> to be closer to those of the original code. I don't know why there is a
> performance regression in Modification 1.
>
> Therefore, I believe modifying the code according to Modification 2 can bring
> maximum benefits. Everyone can discuss whether this approach is acceptable,
> and if so, I will send Patch V4 to proceed with submitting this modification.
>

modification 2 is not correct. if pte0~pte14 are all young and no one
is dirty, we are
having lots of useless "for (; i < CONT_PTES; i++, ptep++)"

                 if (pte_young(pte)) {
                         orig_pte = pte_mkyoung(orig_pte);
                         i++;
                         ptep++;
                         for (; i < CONT_PTES; i++, ptep++) {
                                 pte = __ptep_get(ptep);
                                 if (pte_dirty(pte)) {
                                         orig_pte = pte_mkdirty(orig_pte);
                                         break;
                                 }
                         }
                         break;
                 }

Thanks,
Xavier