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

Xavier
2025-05-01T12:57:42Z

Hi Barry,

At 2025-05-01 07:17:36, "Barry Song" <21cnbao@gmail.com> wrote:
>On Tue, Apr 22, 2025 at 9:34 PM Xavier <xavier_qy@163.com> wrote:
>>
>>
>> Hi all,
>>
>>
>> At 2025-04-16 20:54:47, "Ryan Roberts" <ryan.roberts@arm.com> 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@163.com>
>> >> ---
>> >> 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;
> }
>

I didn't understand which part you referred to when you said there were a lot of
useless loops. According to the scenario you mentioned, "if pte0~pte14 are all
young and no one is dirty", Modification 2 will enter the following branch when
judging pte0:

if (pte_young(pte)) {
orig_pte = pte_mkyoung(orig_pte);
// The dirty status of pte0 has already been checked, skip it.
i++;
ptep++;
// Then we only need to check whether pte1~pte15 are dirty.
for (; i < CONT_PTES; i++, ptep++) {
pte = __ptep_get(ptep);
if (pte_dirty(pte)) {
// Exit as soon as a dirty entry is found.
orig_pte = pte_mkdirty(orig_pte);
break;
}
}
// Exit directly here without going through the outer loop again.
break;
}

In this scenario, the total number of judgments in Modification 2 is nearly half less
than that of the original code. I should have understood it correctly, right?

--

Thanks,
Xavier