core/state/snapshot: implement snapshot layer iteration
This commit is contained in:
Martin Holst Swende
Péter Szilágyi
parent
22c494d399
commit
7e38996301
@ -18,6 +18,7 @@ package snapshot
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import (
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"encoding/binary"
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"bytes"
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"fmt"
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"math"
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"math/rand"
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@ -475,3 +476,291 @@ func (dl *diffLayer) StorageList(accountHash common.Hash) []common.Hash {
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dl.storageList[accountHash] = accountStorageList
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return accountStorageList
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}
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type Iterator interface {
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// Next steps the iterator forward one element, and returns false if
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// the iterator is exhausted
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Next() bool
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// Key returns the current key
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Key() common.Hash
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// Seek steps the iterator forward as many elements as needed, so that after
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// calling Next(), the iterator will be at a key higher than the given hash
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Seek(common.Hash)
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}
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func (dl *diffLayer) newIterator() Iterator {
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dl.AccountList()
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return &dlIterator{dl, -1}
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}
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type dlIterator struct {
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layer *diffLayer
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index int
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}
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func (it *dlIterator) Next() bool {
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if it.index < len(it.layer.accountList) {
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it.index++
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}
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return it.index < len(it.layer.accountList)
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}
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func (it *dlIterator) Key() common.Hash {
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if it.index < len(it.layer.accountList) {
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return it.layer.accountList[it.index]
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}
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return common.Hash{}
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}
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func (it *dlIterator) Seek(key common.Hash) {
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// Search uses binary search to find and return the smallest index i
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// in [0, n) at which f(i) is true
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size := len(it.layer.accountList)
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index := sort.Search(size,
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func(i int) bool {
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v := it.layer.accountList[i]
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return bytes.Compare(key[:], v[:]) < 0
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})
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it.index = index - 1
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}
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type binaryIterator struct {
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a Iterator
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b Iterator
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aDone bool
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bDone bool
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k common.Hash
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}
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func (dl *diffLayer) newBinaryIterator() Iterator {
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parent, ok := dl.parent.(*diffLayer)
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if !ok {
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// parent is the disk layer
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return dl.newIterator()
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}
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l := &binaryIterator{
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a: dl.newIterator(),
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b: parent.newBinaryIterator()}
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l.aDone = !l.a.Next()
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l.bDone = !l.b.Next()
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return l
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}
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func (it *binaryIterator) Next() bool {
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if it.aDone && it.bDone {
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return false
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}
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nextB := it.b.Key()
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first:
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nextA := it.a.Key()
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if it.aDone {
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it.bDone = !it.b.Next()
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it.k = nextB
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return true
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}
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if it.bDone {
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it.aDone = !it.a.Next()
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it.k = nextA
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return true
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}
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if diff := bytes.Compare(nextA[:], nextB[:]); diff < 0 {
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it.aDone = !it.a.Next()
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it.k = nextA
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return true
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} else if diff == 0 {
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// Now we need to advance one of them
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it.aDone = !it.a.Next()
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goto first
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}
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it.bDone = !it.b.Next()
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it.k = nextB
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return true
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}
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func (it *binaryIterator) Key() common.Hash {
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return it.k
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}
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func (it *binaryIterator) Seek(key common.Hash) {
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panic("todo: implement")
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}
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func (dl *diffLayer) iterators() []Iterator {
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if parent, ok := dl.parent.(*diffLayer); ok {
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iterators := parent.iterators()
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return append(iterators, dl.newIterator())
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}
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return []Iterator{dl.newIterator()}
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}
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// fastIterator is a more optimized multi-layer iterator which maintains a
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// direct mapping of all iterators leading down to the bottom layer
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type fastIterator struct {
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iterators []Iterator
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initiated bool
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}
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// Len returns the number of active iterators
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func (fi *fastIterator) Len() int {
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return len(fi.iterators)
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}
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// Less implements sort.Interface
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func (fi *fastIterator) Less(i, j int) bool {
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a := fi.iterators[i].Key()
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b := fi.iterators[j].Key()
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return bytes.Compare(a[:], b[:]) < 0
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}
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// Swap implements sort.Interface
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func (fi *fastIterator) Swap(i, j int) {
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fi.iterators[i], fi.iterators[j] = fi.iterators[j], fi.iterators[i]
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}
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// Next implements the Iterator interface. It returns false if no more elemnts
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// can be retrieved (false == exhausted)
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func (fi *fastIterator) Next() bool {
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if len(fi.iterators) == 0 {
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return false
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}
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if !fi.initiated {
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// Don't forward first time -- we had to 'Next' once in order to
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// do the sorting already
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fi.initiated = true
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return true
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}
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return fi.innerNext(0)
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}
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// innerNext handles the next operation internally,
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// and should be invoked when we know that two elements in the list may have
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// the same value.
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// For example, if the list becomes [2,3,5,5,8,9,10], then we should invoke
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// innerNext(3), which will call Next on elem 3 (the second '5'). It will continue
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// along the list and apply the same operation if needed
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func (fi *fastIterator) innerNext(pos int) bool {
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if !fi.iterators[pos].Next() {
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//Exhausted, remove this iterator
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fi.remove(pos)
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if len(fi.iterators) == 0 {
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return false
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}
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return true
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}
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if pos == len(fi.iterators)-1 {
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// Only one iterator left
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return true
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}
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// We next:ed the elem at 'pos'. Now we may have to re-sort that elem
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val, neighbour := fi.iterators[pos].Key(), fi.iterators[pos+1].Key()
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diff := bytes.Compare(val[:], neighbour[:])
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if diff < 0 {
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// It is still in correct place
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return true
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}
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if diff == 0 {
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// It has same value as the neighbour. So still in correct place, but
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// we need to iterate on the neighbour
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fi.innerNext(pos + 1)
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return true
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}
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// At this point, the elem is in the wrong location, but the
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// remaining list is sorted. Find out where to move the elem
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iterationNeeded := false
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index := sort.Search(len(fi.iterators), func(n int) bool {
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if n <= pos {
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// No need to search 'behind' us
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return false
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}
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if n == len(fi.iterators)-1 {
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// Can always place an elem last
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return true
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}
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neighbour := fi.iterators[n+1].Key()
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diff := bytes.Compare(val[:], neighbour[:])
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if diff == 0 {
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// The elem we're placing it next to has the same value,
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// so it's going to need further iteration
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iterationNeeded = true
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}
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return diff < 0
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})
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fi.move(pos, index)
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if iterationNeeded {
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fi.innerNext(index)
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}
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return true
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}
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// move moves an iterator to another position in the list
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func (fi *fastIterator) move(index, newpos int) {
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if newpos > len(fi.iterators)-1 {
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newpos = len(fi.iterators) - 1
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}
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var (
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elem = fi.iterators[index]
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middle = fi.iterators[index+1 : newpos+1]
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suffix []Iterator
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)
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if newpos < len(fi.iterators)-1 {
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suffix = fi.iterators[newpos+1:]
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}
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fi.iterators = append(fi.iterators[:index], middle...)
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fi.iterators = append(fi.iterators, elem)
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fi.iterators = append(fi.iterators, suffix...)
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}
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// remove drops an iterator from the list
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func (fi *fastIterator) remove(index int) {
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fi.iterators = append(fi.iterators[:index], fi.iterators[index+1:]...)
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}
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// Key returns the current key
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func (fi *fastIterator) Key() common.Hash {
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return fi.iterators[0].Key()
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}
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func (fi *fastIterator) Seek(key common.Hash) {
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// We need to apply this across all iterators
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var seen = make(map[common.Hash]struct{})
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length := len(fi.iterators)
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for i, it := range fi.iterators {
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it.Seek(key)
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for {
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if !it.Next() {
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// To be removed
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// swap it to the last position for now
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fi.iterators[i], fi.iterators[length-1] = fi.iterators[length-1], fi.iterators[i]
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length--
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break
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}
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v := it.Key()
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if _, exist := seen[v]; !exist {
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seen[v] = struct{}{}
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break
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}
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}
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}
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// Now remove those that were placed in the end
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fi.iterators = fi.iterators[:length]
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// The list is now totally unsorted, need to re-sort the entire list
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sort.Sort(fi)
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fi.initiated = false
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}
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// The fast iterator does not query parents as much.
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func (dl *diffLayer) newFastIterator() Iterator {
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f := &fastIterator{dl.iterators(), false}
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f.Seek(common.Hash{})
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return f
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}
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// Debug is a convencience helper during testing
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func (fi *fastIterator) Debug() {
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for _, it := range fi.iterators {
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fmt.Printf(" %v ", it.Key()[31])
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}
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fmt.Println()
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}
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@ -18,6 +18,7 @@ package snapshot
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import (
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"bytes"
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"encoding/binary"
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"math/big"
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"math/rand"
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"testing"
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@ -347,3 +348,365 @@ func BenchmarkJournal(b *testing.B) {
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layer.Journal(new(bytes.Buffer))
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}
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}
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// TestIteratorBasics tests some simple single-layer iteration
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func TestIteratorBasics(t *testing.T) {
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var (
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accounts = make(map[common.Hash][]byte)
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storage = make(map[common.Hash]map[common.Hash][]byte)
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)
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// Fill up a parent
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for i := 0; i < 100; i++ {
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h := randomHash()
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data := randomAccount()
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accounts[h] = data
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if rand.Intn(20) < 10 {
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accStorage := make(map[common.Hash][]byte)
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value := make([]byte, 32)
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rand.Read(value)
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accStorage[randomHash()] = value
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storage[h] = accStorage
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}
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}
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// Add some (identical) layers on top
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parent := newDiffLayer(emptyLayer{}, common.Hash{}, accounts, storage)
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it := parent.newIterator()
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verifyIterator(t, 100, it)
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}
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type testIterator struct {
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values []byte
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}
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func newTestIterator(values ...byte) *testIterator {
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return &testIterator{values}
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}
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func (ti *testIterator) Next() bool {
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ti.values = ti.values[1:]
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if len(ti.values) == 0 {
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return false
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}
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return true
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}
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func (ti *testIterator) Key() common.Hash {
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return common.BytesToHash([]byte{ti.values[0]})
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}
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func (ti *testIterator) Seek(common.Hash) {
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panic("implement me")
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}
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func TestFastIteratorBasics(t *testing.T) {
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type testCase struct {
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lists [][]byte
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expKeys []byte
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}
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for i, tc := range []testCase{
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{lists: [][]byte{{0, 1, 8}, {1, 2, 8}, {2, 9}, {4},
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{7, 14, 15}, {9, 13, 15, 16}},
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expKeys: []byte{0, 1, 2, 4, 7, 8, 9, 13, 14, 15, 16}},
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{lists: [][]byte{{0, 8}, {1, 2, 8}, {7, 14, 15}, {8, 9},
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{9, 10}, {10, 13, 15, 16}},
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expKeys: []byte{0, 1, 2, 7, 8, 9, 10, 13, 14, 15, 16}},
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} {
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var iterators []Iterator
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for _, data := range tc.lists {
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iterators = append(iterators, newTestIterator(data...))
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}
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fi := &fastIterator{
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iterators: iterators,
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initiated: false,
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}
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count := 0
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for fi.Next() {
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if got, exp := fi.Key()[31], tc.expKeys[count]; exp != got {
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t.Errorf("tc %d, [%d]: got %d exp %d", i, count, got, exp)
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}
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count++
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}
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}
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}
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func verifyIterator(t *testing.T, expCount int, it Iterator) {
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var (
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i = 0
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last = common.Hash{}
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)
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for it.Next() {
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v := it.Key()
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if bytes.Compare(last[:], v[:]) >= 0 {
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t.Errorf("Wrong order:\n%x \n>=\n%x", last, v)
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}
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i++
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}
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if i != expCount {
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t.Errorf("iterator len wrong, expected %d, got %d", expCount, i)
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}
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}
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// TestIteratorTraversal tests some simple multi-layer iteration
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func TestIteratorTraversal(t *testing.T) {
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var (
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storage = make(map[common.Hash]map[common.Hash][]byte)
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)
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mkAccounts := func(args ...string) map[common.Hash][]byte {
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accounts := make(map[common.Hash][]byte)
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for _, h := range args {
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accounts[common.HexToHash(h)] = randomAccount()
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}
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return accounts
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}
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// entries in multiple layers should only become output once
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parent := newDiffLayer(emptyLayer{}, common.Hash{},
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mkAccounts("0xaa", "0xee", "0xff", "0xf0"), storage)
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child := parent.Update(common.Hash{},
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mkAccounts("0xbb", "0xdd", "0xf0"), storage)
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child = child.Update(common.Hash{},
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mkAccounts("0xcc", "0xf0", "0xff"), storage)
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// single layer iterator
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verifyIterator(t, 3, child.newIterator())
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// multi-layered binary iterator
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verifyIterator(t, 7, child.newBinaryIterator())
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// multi-layered fast iterator
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verifyIterator(t, 7, child.newFastIterator())
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}
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func TestIteratorLargeTraversal(t *testing.T) {
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// This testcase is a bit notorious -- all layers contain the exact
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// same 200 accounts.
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var storage = make(map[common.Hash]map[common.Hash][]byte)
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mkAccounts := func(num int) map[common.Hash][]byte {
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accounts := make(map[common.Hash][]byte)
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for i := 0; i < num; i++ {
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h := common.Hash{}
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binary.BigEndian.PutUint64(h[:], uint64(i+1))
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accounts[h] = randomAccount()
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}
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return accounts
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}
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parent := newDiffLayer(emptyLayer{}, common.Hash{},
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mkAccounts(200), storage)
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child := parent.Update(common.Hash{},
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mkAccounts(200), storage)
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for i := 2; i < 100; i++ {
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child = child.Update(common.Hash{},
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mkAccounts(200), storage)
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}
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// single layer iterator
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verifyIterator(t, 200, child.newIterator())
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// multi-layered binary iterator
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verifyIterator(t, 200, child.newBinaryIterator())
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// multi-layered fast iterator
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verifyIterator(t, 200, child.newFastIterator())
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}
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// BenchmarkIteratorTraversal is a bit a bit notorious -- all layers contain the exact
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// same 200 accounts. That means that we need to process 2000 items, but only
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// spit out 200 values eventually.
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//
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//BenchmarkIteratorTraversal/binary_iterator-6 2008 573290 ns/op 9520 B/op 199 allocs/op
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//BenchmarkIteratorTraversal/fast_iterator-6 1946 575596 ns/op 20146 B/op 134 allocs/op
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func BenchmarkIteratorTraversal(b *testing.B) {
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var storage = make(map[common.Hash]map[common.Hash][]byte)
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mkAccounts := func(num int) map[common.Hash][]byte {
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accounts := make(map[common.Hash][]byte)
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for i := 0; i < num; i++ {
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h := common.Hash{}
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binary.BigEndian.PutUint64(h[:], uint64(i+1))
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accounts[h] = randomAccount()
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}
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return accounts
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}
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parent := newDiffLayer(emptyLayer{}, common.Hash{},
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mkAccounts(200), storage)
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child := parent.Update(common.Hash{},
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mkAccounts(200), storage)
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for i := 2; i < 100; i++ {
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child = child.Update(common.Hash{},
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mkAccounts(200), storage)
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}
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// We call this once before the benchmark, so the creation of
|
||||
// sorted accountlists are not included in the results.
|
||||
child.newBinaryIterator()
|
||||
b.Run("binary iterator", func(b *testing.B) {
|
||||
for i := 0; i < b.N; i++ {
|
||||
got := 0
|
||||
it := child.newBinaryIterator()
|
||||
for it.Next() {
|
||||
got++
|
||||
}
|
||||
if exp := 200; got != exp {
|
||||
b.Errorf("iterator len wrong, expected %d, got %d", exp, got)
|
||||
}
|
||||
}
|
||||
})
|
||||
b.Run("fast iterator", func(b *testing.B) {
|
||||
for i := 0; i < b.N; i++ {
|
||||
got := 0
|
||||
it := child.newFastIterator()
|
||||
for it.Next() {
|
||||
got++
|
||||
}
|
||||
if exp := 200; got != exp {
|
||||
b.Errorf("iterator len wrong, expected %d, got %d", exp, got)
|
||||
}
|
||||
}
|
||||
})
|
||||
}
|
||||
|
||||
// BenchmarkIteratorLargeBaselayer is a pretty realistic benchmark, where
|
||||
// the baselayer is a lot larger than the upper layer.
|
||||
//
|
||||
// This is heavy on the binary iterator, which in most cases will have to
|
||||
// call recursively 100 times for the majority of the values
|
||||
//
|
||||
// BenchmarkIteratorLargeBaselayer/binary_iterator-6 585 2067377 ns/op 9520 B/op 199 allocs/op
|
||||
// BenchmarkIteratorLargeBaselayer/fast_iterator-6 13198 91043 ns/op 8601 B/op 118 allocs/op
|
||||
func BenchmarkIteratorLargeBaselayer(b *testing.B) {
|
||||
var storage = make(map[common.Hash]map[common.Hash][]byte)
|
||||
|
||||
mkAccounts := func(num int) map[common.Hash][]byte {
|
||||
accounts := make(map[common.Hash][]byte)
|
||||
for i := 0; i < num; i++ {
|
||||
h := common.Hash{}
|
||||
binary.BigEndian.PutUint64(h[:], uint64(i+1))
|
||||
accounts[h] = randomAccount()
|
||||
}
|
||||
return accounts
|
||||
}
|
||||
|
||||
parent := newDiffLayer(emptyLayer{}, common.Hash{},
|
||||
mkAccounts(2000), storage)
|
||||
|
||||
child := parent.Update(common.Hash{},
|
||||
mkAccounts(20), storage)
|
||||
|
||||
for i := 2; i < 100; i++ {
|
||||
child = child.Update(common.Hash{},
|
||||
mkAccounts(20), storage)
|
||||
|
||||
}
|
||||
// We call this once before the benchmark, so the creation of
|
||||
// sorted accountlists are not included in the results.
|
||||
child.newBinaryIterator()
|
||||
b.Run("binary iterator", func(b *testing.B) {
|
||||
for i := 0; i < b.N; i++ {
|
||||
got := 0
|
||||
it := child.newBinaryIterator()
|
||||
for it.Next() {
|
||||
got++
|
||||
}
|
||||
if exp := 2000; got != exp {
|
||||
b.Errorf("iterator len wrong, expected %d, got %d", exp, got)
|
||||
}
|
||||
}
|
||||
})
|
||||
b.Run("fast iterator", func(b *testing.B) {
|
||||
for i := 0; i < b.N; i++ {
|
||||
got := 0
|
||||
it := child.newFastIterator()
|
||||
for it.Next() {
|
||||
got++
|
||||
}
|
||||
if exp := 2000; got != exp {
|
||||
b.Errorf("iterator len wrong, expected %d, got %d", exp, got)
|
||||
}
|
||||
}
|
||||
})
|
||||
}
|
||||
|
||||
// TestIteratorFlatting tests what happens when we
|
||||
// - have a live iterator on child C (parent C1 -> C2 .. CN)
|
||||
// - flattens C2 all the way into CN
|
||||
// - continues iterating
|
||||
// Right now, this "works" simply because the keys do not change -- the
|
||||
// iterator is not aware that a layer has become stale. This naive
|
||||
// solution probably won't work in the long run, however
|
||||
func TestIteratorFlattning(t *testing.T) {
|
||||
var (
|
||||
storage = make(map[common.Hash]map[common.Hash][]byte)
|
||||
)
|
||||
mkAccounts := func(args ...string) map[common.Hash][]byte {
|
||||
accounts := make(map[common.Hash][]byte)
|
||||
for _, h := range args {
|
||||
accounts[common.HexToHash(h)] = randomAccount()
|
||||
}
|
||||
return accounts
|
||||
}
|
||||
// entries in multiple layers should only become output once
|
||||
parent := newDiffLayer(emptyLayer{}, common.Hash{},
|
||||
mkAccounts("0xaa", "0xee", "0xff", "0xf0"), storage)
|
||||
|
||||
child := parent.Update(common.Hash{},
|
||||
mkAccounts("0xbb", "0xdd", "0xf0"), storage)
|
||||
|
||||
child = child.Update(common.Hash{},
|
||||
mkAccounts("0xcc", "0xf0", "0xff"), storage)
|
||||
|
||||
it := child.newFastIterator()
|
||||
child.parent.(*diffLayer).flatten()
|
||||
// The parent should now be stale
|
||||
verifyIterator(t, 7, it)
|
||||
}
|
||||
|
||||
func TestIteratorSeek(t *testing.T) {
|
||||
storage := make(map[common.Hash]map[common.Hash][]byte)
|
||||
mkAccounts := func(args ...string) map[common.Hash][]byte {
|
||||
accounts := make(map[common.Hash][]byte)
|
||||
for _, h := range args {
|
||||
accounts[common.HexToHash(h)] = randomAccount()
|
||||
}
|
||||
return accounts
|
||||
}
|
||||
parent := newDiffLayer(emptyLayer{}, common.Hash{},
|
||||
mkAccounts("0xaa", "0xee", "0xff", "0xf0"), storage)
|
||||
it := parent.newIterator()
|
||||
// expected: ee, f0, ff
|
||||
it.Seek(common.HexToHash("0xdd"))
|
||||
verifyIterator(t, 3, it)
|
||||
|
||||
it = parent.newIterator().(*dlIterator)
|
||||
// expected: ee, f0, ff
|
||||
it.Seek(common.HexToHash("0xaa"))
|
||||
verifyIterator(t, 3, it)
|
||||
|
||||
it = parent.newIterator().(*dlIterator)
|
||||
// expected: nothing
|
||||
it.Seek(common.HexToHash("0xff"))
|
||||
verifyIterator(t, 0, it)
|
||||
|
||||
child := parent.Update(common.Hash{},
|
||||
mkAccounts("0xbb", "0xdd", "0xf0"), storage)
|
||||
|
||||
child = child.Update(common.Hash{},
|
||||
mkAccounts("0xcc", "0xf0", "0xff"), storage)
|
||||
|
||||
it = child.newFastIterator()
|
||||
// expected: cc, dd, ee, f0, ff
|
||||
it.Seek(common.HexToHash("0xbb"))
|
||||
verifyIterator(t, 5, it)
|
||||
|
||||
it = child.newFastIterator()
|
||||
it.Seek(common.HexToHash("0xef"))
|
||||
// exp: f0, ff
|
||||
verifyIterator(t, 2, it)
|
||||
|
||||
it = child.newFastIterator()
|
||||
it.Seek(common.HexToHash("0xf0"))
|
||||
verifyIterator(t, 1, it)
|
||||
|
||||
it.Seek(common.HexToHash("0xff"))
|
||||
verifyIterator(t, 0, it)
|
||||
|
||||
}
|
||||
|
||||
60
core/state/snapshot/iteration.md
Normal file
60
core/state/snapshot/iteration.md
Normal file
@ -0,0 +1,60 @@
|
||||
|
||||
## How the fast iterator works
|
||||
|
||||
Consider the following example, where we have `6` iterators, sorted from
|
||||
left to right in ascending order.
|
||||
|
||||
Our 'primary' `A` iterator is on the left, containing the elements `[0,1,8]`
|
||||
```
|
||||
A B C D E F
|
||||
|
||||
0 1 2 4 7 9
|
||||
1 2 9 - 14 13
|
||||
8 8 - 15 15
|
||||
- - - 16
|
||||
-
|
||||
```
|
||||
When we call `Next` on the primary iterator, we get (ignoring the future keys)
|
||||
|
||||
```
|
||||
A B C D E F
|
||||
|
||||
1 1 2 4 7 9
|
||||
```
|
||||
We detect that we now got an equality between our element and the next element.
|
||||
And we need to continue `Next`ing on the next element
|
||||
|
||||
```
|
||||
1 2 2 4 7 9
|
||||
```
|
||||
And move on:
|
||||
```
|
||||
A B C D E F
|
||||
|
||||
1 2 9 4 7 9
|
||||
```
|
||||
Now we broke out of the equality, but we need to re-sort the element `C`
|
||||
|
||||
```
|
||||
A B D E F C
|
||||
|
||||
1 2 4 7 9 9
|
||||
```
|
||||
|
||||
And after shifting it rightwards, we check equality again, and find `C == F`, and thus
|
||||
call `Next` on `C`
|
||||
|
||||
```
|
||||
A B D E F C
|
||||
|
||||
1 2 4 7 9 -
|
||||
```
|
||||
At this point, `C` was exhausted, and is removed
|
||||
|
||||
```
|
||||
A B D E F
|
||||
|
||||
1 2 4 7 9
|
||||
```
|
||||
And we're done with this step.
|
||||
|
||||
Loading…
Reference in New Issue
Block a user