The size of struct sched_domain_topology_level is 64 bytes.
For NUMA platforms, almost all are multi-core (enable CONFIG_SCHED_MC),
That is to say, the default_topology array has at least 128 bytes that
need to be copied in sched_init_numa(). For most x86 platforms,
CONFIG_SCHED_SMT will be enabled, so more copies will be required.
And memcpy() will be optimized under different architectures.
Fortunately, for platforms with CONFIG_NUMA enabled,
these optimizations are likely to be used.
So, let's use memcpy to copy default topology in sched_init_numa().
Tests are done in an Intel Xeon(R) Platinum 8260 [email protected] machine
with 2 NUMA nodes each of which has 24 cores with SMT2 enabled, so 96
CPUs in total.
Use RDTSC to count time-consuming, and based on 5.19-rc4.
Enable CONFIG_SCHED_SMT && CONFIG_SCHED_CLUSTER && CONFIG_SCHED_MC,
So the default_topology array has 256 bytes that need to be copied
in sched_init_numa().
5.19-rc4 5.19-rc4 with patch
average tsc ticks 516.57 85.33 （-83.48%*）
Enable CONFIG_SCHED_MC, So the default_topology array has
128 bytes that need to be copied in sched_init_numa().
5.19-rc4 5.19-rc4 with patch
average tsc ticks 65.71 55.00 （-16.30%*）
Signed-off-by: Hao Jia <[email protected]>
---
kernel/sched/topology.c | 3 +--
1 file changed, 1 insertion(+), 2 deletions(-)
diff --git a/kernel/sched/topology.c b/kernel/sched/topology.c
index 05b6c2ad90b9..c6f497d263cd 100644
--- a/kernel/sched/topology.c
+++ b/kernel/sched/topology.c
@@ -1918,8 +1918,7 @@ void sched_init_numa(int offline_node)
/*
* Copy the default topology bits..
*/
- for (i = 0; sched_domain_topology[i].mask; i++)
- tl[i] = sched_domain_topology[i];
+ memcpy(tl, sched_domain_topology, sizeof(struct sched_domain_topology_level) * i);
/*
* Add the NUMA identity distance, aka single NODE.
--
2.32.0
On 27/06/22 18:53, Hao Jia wrote:
> The size of struct sched_domain_topology_level is 64 bytes.
> For NUMA platforms, almost all are multi-core (enable CONFIG_SCHED_MC),
> That is to say, the default_topology array has at least 128 bytes that
> need to be copied in sched_init_numa(). For most x86 platforms,
> CONFIG_SCHED_SMT will be enabled, so more copies will be required.
>
> And memcpy() will be optimized under different architectures.
> Fortunately, for platforms with CONFIG_NUMA enabled,
> these optimizations are likely to be used.
> So, let's use memcpy to copy default topology in sched_init_numa().
>
> Tests are done in an Intel Xeon(R) Platinum 8260 [email protected] machine
> with 2 NUMA nodes each of which has 24 cores with SMT2 enabled, so 96
> CPUs in total.
>
> Use RDTSC to count time-consuming, and based on 5.19-rc4.
>
> Enable CONFIG_SCHED_SMT && CONFIG_SCHED_CLUSTER && CONFIG_SCHED_MC,
> So the default_topology array has 256 bytes that need to be copied
> in sched_init_numa().
> 5.19-rc4 5.19-rc4 with patch
> average tsc ticks 516.57 85.33 （-83.48%*）
>
> Enable CONFIG_SCHED_MC, So the default_topology array has
> 128 bytes that need to be copied in sched_init_numa().
> 5.19-rc4 5.19-rc4 with patch
> average tsc ticks 65.71 55.00 （-16.30%*）
>
> Signed-off-by: Hao Jia <[email protected]>
It's not a very hot path but I guess this lets you shave off a bit of boot
time... While you're at it, you could add an early
if (nr_node_ids == 1)
return;
since !NUMA systems still go through sched_init_numa() if they have a
kernel with CONFIG_NUMA (which should be most of them nowdays) and IIRC
they end up with an unused NODE topology level.
Regardless:
Reviewed-by: Valentin Schneider <[email protected]>
On 2022/7/4 Valentin Schneider wrote:
>
> It's not a very hot path but I guess this lets you shave off a bit of boot
> time... While you're at it, you could add an early
Thanks for your time and suggestion.
>
> if (nr_node_ids == 1)
> return;
>
This will cause the values of sched_domains_numa_levels and
sched_max_numa_distance to be different from before, and
sched_domains_numa_levels may cause the return value of
sched_numa_find_closest() to be different.
I'm not sure if it will cause problems.
> since !NUMA systems still go through sched_init_numa() if they have a
> kernel with CONFIG_NUMA (which should be most of them nowdays) and IIRC
> they end up with an unused NODE topology level.
>
I'm confused why most !NUMA systems enable CONFIG_NUMA in the kernel?
Maybe for scalability?
> Regardless:
>
> Reviewed-by: Valentin Schneider <[email protected]>
>
How about this?
The size of struct sched_domain_topology_level is 64 bytes.
For NUMA platforms, almost all are multi-core (enable CONFIG_SCHED_MC),
That is to say, the default_topology array has at least 128 bytes that
need to be copied in sched_init_numa(). For most x86 platforms,
CONFIG_SCHED_SMT will be enabled, so more copies will be required.
And memcpy() will be optimized under different architectures.
Fortunately, for platforms with CONFIG_NUMA enabled,
these optimizations are likely to be used.
So, let's use memcpy() to copy default topology in sched_init_numa().
Tests are done in an Intel Xeon(R) Platinum 8260 [email protected] machine
with 2 NUMA nodes each of which has 24 cores with SMT2 enabled, so 96
CPUs in total.
Use RDTSC to count time-consuming, and based on 5.19-rc4.
Enable CONFIG_SCHED_SMT && CONFIG_SCHED_CLUSTER && CONFIG_SCHED_MC,
So the default_topology array has 256 bytes that need to be copied
in sched_init_numa().
5.19-rc4 5.19-rc4 with patch
average tsc ticks 516.57 85.33 （-83.48%*）
Enable CONFIG_SCHED_MC, So the default_topology array has
128 bytes that need to be copied in sched_init_numa().
5.19-rc4 5.19-rc4 with patch
average tsc ticks 65.71 55.00 （-16.30%*）
since !NUMA systems still go through sched_init_numa() if they have a
kernel with CONFIG_NUMA (which should be most of them nowdays) and we
can skip copying and memory allocation of useless default topology.
Suggested-by: Valentin Schneider <[email protected]>
Signed-off-by: Hao Jia <[email protected]>
---
kernel/sched/topology.c | 7 +++++--
1 file changed, 5 insertions(+), 2 deletions(-)
diff --git a/kernel/sched/topology.c b/kernel/sched/topology.c
index 05b6c2ad90b9..c439e58f22b9 100644
--- a/kernel/sched/topology.c
+++ b/kernel/sched/topology.c
@@ -1907,6 +1907,9 @@ void sched_init_numa(int offline_node)
}
rcu_assign_pointer(sched_domains_numa_masks, masks);
+ if (nr_node_ids == 1)
+ goto skip;
+
/* Compute default topology size */
for (i = 0; sched_domain_topology[i].mask; i++);
@@ -1918,8 +1921,7 @@ void sched_init_numa(int offline_node)
/*
* Copy the default topology bits..
*/
- for (i = 0; sched_domain_topology[i].mask; i++)
- tl[i] = sched_domain_topology[i];
+ memcpy(tl, sched_domain_topology, sizeof(struct
sched_domain_topology_level) * i);
/*
* Add the NUMA identity distance, aka single NODE.
@@ -1946,6 +1948,7 @@ void sched_init_numa(int offline_node)
sched_domain_topology_saved = sched_domain_topology;
sched_domain_topology = tl;
+skip:
sched_domains_numa_levels = nr_levels;
WRITE_ONCE(sched_max_numa_distance,
sched_domains_numa_distance[nr_levels - 1]);
thanks,
Hao
On 11/07/22 18:28, Hao Jia wrote:
> On 2022/7/4 Valentin Schneider wrote:
>>
>> It's not a very hot path but I guess this lets you shave off a bit of boot
>> time... While you're at it, you could add an early
> Thanks for your time and suggestion.
>>
>> if (nr_node_ids == 1)
>> return;
>>
>
> This will cause the values of sched_domains_numa_levels and
> sched_max_numa_distance to be different from before, and
> sched_domains_numa_levels may cause the return value of
> sched_numa_find_closest() to be different.
> I'm not sure if it will cause problems.
>
True, we need to be careful here, but those are all static so they get
initialized to sensible defaults (zero / NULL pointer).
sched_numa_find_closest() will return nr_cpu_ids which make sense, so I
think we can get away with an early return
>> since !NUMA systems still go through sched_init_numa() if they have a
>> kernel with CONFIG_NUMA (which should be most of them nowdays) and IIRC
>> they end up with an unused NODE topology level.
>>
>
> I'm confused why most !NUMA systems enable CONFIG_NUMA in the kernel?
> Maybe for scalability?
>
It just makes things easier on a distribution point of view - just ship a
single kernel image everyone can use, rather than N different images for N
different types of systems.
AFAIA having CONFIG_NUMA on an UMA (!NUMA) system isn't bad, it just adds
more things in the sched_domain_topology during boot time which end up
being unused.
On 2022/7/12 Valentin Schneider wrote:
> On 11/07/22 18:28, Hao Jia wrote:
>> On 2022/7/4 Valentin Schneider wrote:
>>>
>>> It's not a very hot path but I guess this lets you shave off a bit of boot
>>> time... While you're at it, you could add an early
>> Thanks for your time and suggestion.
>>>
>>> if (nr_node_ids == 1)
>>> return;
>>>
>>
>> This will cause the values of sched_domains_numa_levels and
>> sched_max_numa_distance to be different from before, and
>> sched_domains_numa_levels may cause the return value of
>> sched_numa_find_closest() to be different.
>> I'm not sure if it will cause problems.
>>
>
> True, we need to be careful here, but those are all static so they get
> initialized to sensible defaults (zero / NULL pointer).
>
> sched_numa_find_closest() will return nr_cpu_ids which make sense, so I
> think we can get away with an early return
Yes, this may affect the return value of housekeeping anycpu(), which
doesn't seem to be a problem.
>
>>> since !NUMA systems still go through sched_init_numa() if they have a
>>> kernel with CONFIG_NUMA (which should be most of them nowdays) and IIRC
>>> they end up with an unused NODE topology level.
>>>
>>
>> I'm confused why most !NUMA systems enable CONFIG_NUMA in the kernel?
>> Maybe for scalability?
>>
>
> It just makes things easier on a distribution point of view - just ship a
> single kernel image everyone can use, rather than N different images for N
> different types of systems.
>
> AFAIA having CONFIG_NUMA on an UMA (!NUMA) system isn't bad, it just adds
> more things in the sched_domain_topology during boot time which end up
> being unused.
Thank you very much for your answer.
Thinks,
Hao
>