It looks the _PageSize constant used in allocating large memory block is
declared as 8192 in code, but os.Getpagesize() returns 4096.
const (
...
_PageShift = 13
_PageSize = 1 << _PageShift
...
)
Is this intended?
On Friday, June 11, 2021 at 7:59:27 PM UTC-4 tapi...@gmail.com wrote:
> On Friday, June 11, 2021 at 2:00:03 PM UTC-4 Ian Lance Taylor wrote:
>
>> On Fri, Jun 11, 2021 at 9:38 AM tapi...@gmail.com <tapi...@gmail.com>
>> wrote:
>> >
>> > package allocate
>> >
>> > import "testing"
>> > import "os"
>> > import "fmt"
>> >
>> > func init() {
>> > fmt.Println("OS page size:", os.Getpagesize())
>> > }
>> >
>> > var r1 []byte
>> >
>> > func BenchmarkCount1(b *testing.B) {
>> > for i := 0; i < b.N; i++ {
>> > r1 = make([]byte, 32768+1)
>> > }
>> > }
>> >
>> > var r2 []byte
>> >
>> > func BenchmarkCount2(b *testing.B) {
>> > for i := 0; i < b.N; i++ {
>> > r2 = make([]byte, 40)
>> > }
>> > }
>> >
>> > The output:
>> >
>> > OS page size: 32768
>> > BenchmarkCount1-4 166443 7312 ns/op 40960 B/op 1 allocs/op
>> > BenchmarkCount2-4 31465389 36.88 ns/op 48 B/op 1 allocs/op
>> >
>> > In fact, a memory block with size 36864 is enough to carry the elements
>> of a byte slice with size 32768+1. Why to allocate one more page for them?
>>
>> Memory blocks larger than 32768 bytes flip over to the "large
>> allocation" model. See _MaxSmallSize in runtime/sizeclasses.go (a
>> generated file). This requires an extra span. I haven't tried to
>> work it out in detail, but it probably has something to do with that.
>>
>> Ian
>>
>
> Yes, that is why I chose a size 32768+1.
> I understand that the large memory block will be a multiple of page size.
> What I haven't get is why one more page is allocated.
>
> I will try to read the source to get the answer.
>
> BTW, sorry, I missed a "more" in the thread title.
>
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