sync.Pool unexpected behavior - go

Question

Getting to know the sync.Pool i wrote 2 different New methods for it:
Returns a slice;
Returns a ptr to slice;
Just alloc a new slice without pool (for compare performance);
Benchmarks surprised me
6ns/op and 1 alloc
17ns/op and 0 alloc
130ns/op and 1 alloc
Please explain why is such a big difference between 1 and 3, given that both allocate memory?
And why 1st allocates?
package main
import (
"bytes"
"sync"
)
func main() {}
type MyBuffer struct {
pool sync.Pool
}
func New() *MyBuffer {
return &MyBuffer{pool: sync.Pool{
New: func() any {
buf := make([]int, 0, 128)
return &buf
},
}}
}
func New2() *MyBuffer {
return &MyBuffer{pool: sync.Pool{
New: func() any {
buf := make([]int, 0, 128)
return buf
},
}}
}
func Fill[S ~[]E, E any](buf *S) {
var some E
for i := 0; i < cap(*buf); i++ {
*buf = append(*buf, some)
}
}
func BenchmarkAlloc(b *testing.B) {
b.RunParallel(func(p *testing.PB) {
for p.Next() {
// Get buf
buf := make([]int, 0, 128)
// fill
Fill(&buf)
// Reset
buf = buf[:0]
}
})
}
func BenchmarkPoolPtr(b *testing.B) {
mb := New()
b.RunParallel(func(p *testing.PB) {
for p.Next() {
// Get buf
buf, _ := mb.pool.Get().(*[]int)
// fill
Fill(buf)
// reset
*buf = (*buf)[:0]
// put
mb.pool.Put(buf)
}
})
}
func BenchmarkPoolNoPtr(b *testing.B) {
mb := New2()
b.RunParallel(func(p *testing.PB) {
for p.Next() {
// Get buf
buf, _ := mb.pool.Get().([]int)
// fill
Fill(&buf)
// reset
buf = buf[:0]
// put
mb.pool.Put(&buf)
}
})
}

Answer 1

I think in the 1st case, the allocation is a reference to your slice. In the 3rd case, the allocation is the complete slice.
When you look the size allocated, it's 32 B/op vs 1024 B/op
so the allocation is the reference to your slice, that is not needed when your variable is a pointer.