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sync: implement simple pooling in sync.Pool

pull/3269/head
Anuraag Agrawal 2 years ago
committed by Ron Evans
parent
commit
29f8d22a2d
  1. 12
      src/sync/pool.go
  2. 50
      src/sync/pool_test.go

12
src/sync/pool.go

@ -1,13 +1,18 @@
package sync
// Pool is a very simple implementation of sync.Pool. It does not actually
// implement a pool.
// Pool is a very simple implementation of sync.Pool.
type Pool struct {
New func() interface{}
New func() interface{}
items []interface{}
}
// Get returns the value of calling Pool.New().
func (p *Pool) Get() interface{} {
if len(p.items) > 0 {
x := p.items[len(p.items)-1]
p.items = p.items[:len(p.items)-1]
return x
}
if p.New == nil {
return nil
}
@ -16,4 +21,5 @@ func (p *Pool) Get() interface{} {
// Put drops the value put into the pool.
func (p *Pool) Put(x interface{}) {
p.items = append(p.items, x)
}

50
src/sync/pool_test.go

@ -0,0 +1,50 @@
package sync_test
import (
"sync"
"testing"
)
type testItem struct {
val int
}
func TestPool(t *testing.T) {
p := sync.Pool{
New: func() interface{} {
return &testItem{}
},
}
i1P := p.Get()
if i1P == nil {
t.Error("pool with New returned nil")
}
i1 := i1P.(*testItem)
if got, want := i1.val, 0; got != want {
t.Errorf("empty pool item value: got %v, want %v", got, want)
}
i1.val = 1
i2 := p.Get().(*testItem)
if got, want := i2.val, 0; got != want {
t.Errorf("empty pool item value: got %v, want %v", got, want)
}
i2.val = 2
p.Put(i1)
i3 := p.Get().(*testItem)
if got, want := i3.val, 1; got != want {
t.Errorf("pool with item value: got %v, want %v", got, want)
}
}
func TestPool_noNew(t *testing.T) {
p := sync.Pool{}
i1 := p.Get()
if i1 != nil {
t.Errorf("pool without New returned %v, want nil", i1)
}
}
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