Evgenii Stratonikov
15102e6dfd
`slices.SortFunc` doesn't use reflection and is a bit faster. I have done some micro-benchmarks for `[]NodeInfo`: ``` $ benchstat -col "/func" out goos: linux goarch: amd64 pkg: git.frostfs.info/TrueCloudLab/frostfs-node/pkg/local_object_storage/pilorama cpu: 11th Gen Intel(R) Core(TM) i5-1135G7 @ 2.40GHz │ sort.Slice │ slices.SortFunc │ │ sec/op │ sec/op vs base │ Sort-8 2.130µ ± 2% 1.253µ ± 2% -41.20% (p=0.000 n=10) ``` Haven't included them, though, as they I don't see them being used a lot. Signed-off-by: Evgenii Stratonikov <e.stratonikov@yadro.com>
118 lines
3.2 KiB
Go
118 lines
3.2 KiB
Go
package pilorama
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import (
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"cmp"
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"encoding/binary"
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"slices"
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"sync"
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"time"
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cidSDK "git.frostfs.info/TrueCloudLab/frostfs-sdk-go/container/id"
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"go.etcd.io/bbolt"
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)
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type batch struct {
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forest *boltForest
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timer *time.Timer
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// mtx protects timer and operations fields.
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// Because mtx can be taken inside a transaction,
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// transactions MUST NOT be executed with the mutex taken to avoid a deadlock.
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mtx sync.Mutex
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start sync.Once
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cid cidSDK.ID
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treeID string
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results []chan<- error
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operations []*Move
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}
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func (b *batch) trigger() {
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b.mtx.Lock()
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if b.timer != nil {
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b.timer.Stop()
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}
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b.mtx.Unlock()
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b.start.Do(b.run)
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}
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func (b *batch) run() {
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fullID := bucketName(b.cid, b.treeID)
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err := b.forest.db.Update(func(tx *bbolt.Tx) error {
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bLog, bTree, err := b.forest.getTreeBuckets(tx, fullID)
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if err != nil {
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return err
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}
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b.mtx.Lock()
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b.timer = nil
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b.mtx.Unlock()
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// Sorting without a mutex is ok, because we append to this slice only if timer is non-nil.
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// See (*boltForest).addBatch for details.
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slices.SortFunc(b.operations, func(mi, mj *Move) int {
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return cmp.Compare(mi.Time, mj.Time)
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})
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b.operations = slices.CompactFunc(b.operations, func(x, y *Move) bool { return x.Time == y.Time })
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// Our main use-case is addition of new items. In this case,
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// we do not need to perform undo()/redo(), just do().
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// https://github.com/trvedata/move-op/blob/6c23447c12a7862ff31b7fc2205f6c90fbdb9dc0/proof/Move_Create.thy#L259
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//
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// For this optimization to work we need to ensure three things:
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// 1. The node itself is not yet in tree.
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// 2. The node is not a parent. This case is not mentioned in the article, because
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// they consider a "static order" (perform all CREATE operations before MOVE).
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// We need this because if node _is_ a parent, we could violate (3) for some late operation.
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// See TestForest_ApplySameOperation for details.
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// 3. Parent of each operation is already in tree.
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var parents map[uint64]struct{}
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var cKey [maxKeySize]byte
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var slow bool
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for i := range b.operations {
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_, _, _, inTree := b.forest.getState(bTree, stateKey(cKey[:], b.operations[i].Child))
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if inTree {
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slow = true
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break
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}
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key := childrenKey(cKey[:], b.operations[i].Child, 0)
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k, _ := bTree.Cursor().Seek(key)
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if len(k) == childrenKeySize && binary.LittleEndian.Uint64(k[1:]) == b.operations[i].Child {
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slow = true
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break
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}
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if b.operations[i].Parent == RootID {
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continue
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} else if parents == nil {
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// Attaching key only to root is done frequently,
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// no allocations are performed unless necessary.
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parents = make(map[uint64]struct{})
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} else if _, ok := parents[b.operations[i].Parent]; ok {
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continue
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}
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p := b.operations[i].Parent
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_, ts, _, inTree := b.forest.getState(bTree, stateKey(cKey[:], p))
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if !inTree || b.operations[0].Time < ts {
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slow = true
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break
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}
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parents[b.operations[i].Parent] = struct{}{}
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}
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if slow {
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var lm Move
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return b.forest.applyOperation(bLog, bTree, b.operations, &lm)
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}
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for i := range b.operations {
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if err := b.forest.do(bLog, bTree, cKey[:], b.operations[i]); err != nil {
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return err
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}
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}
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return nil
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})
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for i := range b.results {
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b.results[i] <- err
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}
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}
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