On Fri, Jun 11, 2021 at 9:38 AM tapi...@gmail.com <tapir....@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
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