Files
mesh-controller/vendor/github.com/nats-io/nats-server/v2/server/stree/stree.go
T

511 lines
15 KiB
Go

// Copyright 2023-2025 The NATS Authors
// Licensed under the Apache License, Version 2.0 (the "License");
// you may not use this file except in compliance with the License.
// You may obtain a copy of the License at
//
// http://www.apache.org/licenses/LICENSE-2.0
//
// Unless required by applicable law or agreed to in writing, software
// distributed under the License is distributed on an "AS IS" BASIS,
// WITHOUT WARRANTIES OR CONDITIONS OF ANY KIND, either express or implied.
// See the License for the specific language governing permissions and
// limitations under the License.
package stree
import (
"bytes"
"slices"
"unsafe"
"github.com/nats-io/nats-server/v2/server/gsl"
)
// SubjectTree is an adaptive radix trie (ART) for storing subject information on literal subjects.
// Will use dynamic nodes, path compression and lazy expansion.
// The reason this exists is to not only save some memory in our filestore but to greatly optimize matching
// a wildcard subject to certain members, e.g. consumer NumPending calculations.
type SubjectTree[T any] struct {
root node
size int
}
// NewSubjectTree creates a new SubjectTree with values T.
func NewSubjectTree[T any]() *SubjectTree[T] {
return &SubjectTree[T]{}
}
// Size returns the number of elements stored.
func (t *SubjectTree[T]) Size() int {
if t == nil {
return 0
}
return t.size
}
// Will empty out the tree, or if tree is nil create a new one.
func (t *SubjectTree[T]) Empty() *SubjectTree[T] {
if t == nil {
return NewSubjectTree[T]()
}
t.root, t.size = nil, 0
return t
}
// Insert a value into the tree. Will return if the value was updated and if so the old value.
func (t *SubjectTree[T]) Insert(subject []byte, value T) (*T, bool) {
if t == nil {
return nil, false
}
// Make sure we never insert anything with a noPivot byte.
if bytes.IndexByte(subject, noPivot) >= 0 {
return nil, false
}
old, updated := t.insert(&t.root, subject, value, 0)
if !updated {
t.size++
}
return old, updated
}
// Find will find the value and return it or false if it was not found.
func (t *SubjectTree[T]) Find(subject []byte) (*T, bool) {
if t == nil {
return nil, false
}
var si int
for n := t.root; n != nil; {
if n.isLeaf() {
if ln := n.(*leaf[T]); ln.match(subject[si:]) {
return &ln.value, true
}
return nil, false
}
// We are a node type here, grab meta portion.
if bn := n.base(); len(bn.prefix) > 0 {
end := min(si+len(bn.prefix), len(subject))
if !bytes.Equal(subject[si:end], bn.prefix) {
return nil, false
}
// Increment our subject index.
si += len(bn.prefix)
}
if an := n.findChild(pivot(subject, si)); an != nil {
n = *an
} else {
return nil, false
}
}
return nil, false
}
// Delete will delete the item and return its value, or not found if it did not exist.
func (t *SubjectTree[T]) Delete(subject []byte) (*T, bool) {
if t == nil {
return nil, false
}
val, deleted := t.delete(&t.root, subject, 0)
if deleted {
t.size--
}
return val, deleted
}
// Match will match against a subject that can have wildcards and invoke the callback func for each matched value.
func (t *SubjectTree[T]) Match(filter []byte, cb func(subject []byte, val *T)) {
if t == nil || t.root == nil || len(filter) == 0 || cb == nil {
return
}
// We need to break this up into chunks based on wildcards, either pwc '*' or fwc '>'.
var raw [16][]byte
parts := genParts(filter, raw[:0])
var _pre [256]byte
t.match(t.root, parts, _pre[:0], cb)
}
// IterOrdered will walk all entries in the SubjectTree lexicographically. The callback can return false to terminate the walk.
func (t *SubjectTree[T]) IterOrdered(cb func(subject []byte, val *T) bool) {
if t == nil || t.root == nil {
return
}
var _pre [256]byte
t.iter(t.root, _pre[:0], true, cb)
}
// IterFast will walk all entries in the SubjectTree with no guarantees of ordering. The callback can return false to terminate the walk.
func (t *SubjectTree[T]) IterFast(cb func(subject []byte, val *T) bool) {
if t == nil || t.root == nil {
return
}
var _pre [256]byte
t.iter(t.root, _pre[:0], false, cb)
}
// Internal methods
// Internal call to insert that can be recursive.
func (t *SubjectTree[T]) insert(np *node, subject []byte, value T, si int) (*T, bool) {
n := *np
if n == nil {
*np = newLeaf(subject, value)
return nil, false
}
if n.isLeaf() {
ln := n.(*leaf[T])
if ln.match(subject[si:]) {
// Replace with new value.
old := ln.value
ln.value = value
return &old, true
}
// Here we need to split this leaf.
cpi := commonPrefixLen(ln.suffix, subject[si:])
nn := newNode4(subject[si : si+cpi])
ln.setSuffix(ln.suffix[cpi:])
si += cpi
// Make sure we have different pivot, normally this will be the case unless we have overflowing prefixes.
if p := pivot(ln.suffix, 0); cpi > 0 && si < len(subject) && p == subject[si] {
// We need to split the original leaf. Recursively call into insert.
t.insert(np, subject, value, si)
// Now add the update version of *np as a child to the new node4.
nn.addChild(p, *np)
} else {
// Can just add this new leaf as a sibling.
nl := newLeaf(subject[si:], value)
nn.addChild(pivot(nl.suffix, 0), nl)
// Add back original.
nn.addChild(pivot(ln.suffix, 0), ln)
}
*np = nn
return nil, false
}
// Non-leaf nodes.
bn := n.base()
if len(bn.prefix) > 0 {
cpi := commonPrefixLen(bn.prefix, subject[si:])
if pli := len(bn.prefix); cpi >= pli {
// Move past this node. We look for an existing child node to recurse into.
// If one does not exist we can create a new leaf node.
si += pli
if nn := n.findChild(pivot(subject, si)); nn != nil {
return t.insert(nn, subject, value, si)
}
if n.isFull() {
n = n.grow()
*np = n
}
n.addChild(pivot(subject, si), newLeaf(subject[si:], value))
return nil, false
} else {
// We did not match the prefix completely here.
// Calculate new prefix for this node.
prefix := subject[si : si+cpi]
si += len(prefix)
// We will insert a new node4 and attach our current node below after adjusting prefix.
nn := newNode4(prefix)
// Shift the prefix for our original node.
n.setPrefix(bn.prefix[cpi:])
nn.addChild(pivot(bn.prefix[:], 0), n)
// Add in our new leaf.
nn.addChild(pivot(subject[si:], 0), newLeaf(subject[si:], value))
// Update our node reference.
*np = nn
}
} else {
if nn := n.findChild(pivot(subject, si)); nn != nil {
return t.insert(nn, subject, value, si)
}
// No prefix and no matched child, so add in new leafnode as needed.
if n.isFull() {
n = n.grow()
*np = n
}
n.addChild(pivot(subject, si), newLeaf(subject[si:], value))
}
return nil, false
}
// internal function to recursively find the leaf to delete. Will do compaction if the item is found and removed.
func (t *SubjectTree[T]) delete(np *node, subject []byte, si int) (*T, bool) {
if t == nil || np == nil || *np == nil || len(subject) == 0 {
return nil, false
}
n := *np
if n.isLeaf() {
ln := n.(*leaf[T])
if ln.match(subject[si:]) {
*np = nil
return &ln.value, true
}
return nil, false
}
// Not a leaf node.
if bn := n.base(); len(bn.prefix) > 0 {
// subject could be shorter and would panic on bad index into subject slice.
if len(subject) < si+len(bn.prefix) {
return nil, false
}
if !bytes.Equal(subject[si:si+len(bn.prefix)], bn.prefix) {
return nil, false
}
// Increment our subject index.
si += len(bn.prefix)
}
p := pivot(subject, si)
nna := n.findChild(p)
if nna == nil {
return nil, false
}
nn := *nna
if nn.isLeaf() {
ln := nn.(*leaf[T])
if ln.match(subject[si:]) {
n.deleteChild(p)
if sn := n.shrink(); sn != nil {
bn := n.base()
// Make sure to set cap so we force an append to copy below.
pre := bn.prefix[:len(bn.prefix):len(bn.prefix)]
// Need to fix up prefixes/suffixes.
if sn.isLeaf() {
ln := sn.(*leaf[T])
// Make sure to set cap so we force an append to copy.
ln.suffix = append(pre, ln.suffix...)
} else {
// We are a node here, we need to add in the old prefix.
if len(pre) > 0 {
bsn := sn.base()
sn.setPrefix(append(pre, bsn.prefix...))
}
}
*np = sn
}
return &ln.value, true
}
return nil, false
}
return t.delete(nna, subject, si)
}
// Internal function which can be called recursively to match all leaf nodes to a given filter subject which
// once here has been decomposed to parts. These parts only care about wildcards, both pwc and fwc.
func (t *SubjectTree[T]) match(n node, parts [][]byte, pre []byte, cb func(subject []byte, val *T)) {
// Capture if we are sitting on a terminal fwc.
var hasFWC bool
if lp := len(parts); lp > 0 && len(parts[lp-1]) > 0 && parts[lp-1][0] == fwc {
hasFWC = true
}
for n != nil {
nparts, matched := n.matchParts(parts)
// Check if we did not match.
if !matched {
return
}
// We have matched here. If we are a leaf and have exhausted all parts or he have a FWC fire callback.
if n.isLeaf() {
if len(nparts) == 0 || (hasFWC && len(nparts) == 1) {
ln := n.(*leaf[T])
cb(append(pre, ln.suffix...), &ln.value)
}
return
}
// We have normal nodes here.
// We need to append our prefix
bn := n.base()
if len(bn.prefix) > 0 {
// Note that this append may reallocate, but it doesn't modify "pre" at the "match" callsite.
pre = append(pre, bn.prefix...)
}
// Check our remaining parts.
if len(nparts) == 0 && !hasFWC {
// We are a node with no parts left and we are not looking at a fwc.
// We could have a leafnode with no suffix which would be a match.
// We could also have a terminal pwc. Check for those here.
var hasTermPWC bool
if lp := len(parts); lp > 0 && len(parts[lp-1]) == 1 && parts[lp-1][0] == pwc {
// If we are sitting on a terminal pwc, put the pwc back and continue.
nparts = parts[len(parts)-1:]
hasTermPWC = true
}
for _, cn := range n.children() {
if cn == nil {
continue
}
if cn.isLeaf() {
ln := cn.(*leaf[T])
if len(ln.suffix) == 0 {
cb(append(pre, ln.suffix...), &ln.value)
} else if hasTermPWC && bytes.IndexByte(ln.suffix, tsep) < 0 {
cb(append(pre, ln.suffix...), &ln.value)
}
} else if hasTermPWC {
// We have terminal pwc so call into match again with the child node.
t.match(cn, nparts, pre, cb)
}
}
// Return regardless.
return
}
// If we are sitting on a terminal fwc, put back and continue.
if hasFWC && len(nparts) == 0 {
nparts = parts[len(parts)-1:]
}
// Here we are a node type with a partial match.
// Check if the first part is a wildcard.
fp := nparts[0]
p := pivot(fp, 0)
// Check if we have a pwc/fwc part here. This will cause us to iterate.
if len(fp) == 1 && (p == pwc || p == fwc) {
// We need to iterate over all children here for the current node
// to see if we match further down.
for _, cn := range n.children() {
if cn != nil {
t.match(cn, nparts, pre, cb)
}
}
return
}
// Here we have normal traversal, so find the next child.
nn := n.findChild(p)
if nn == nil {
return
}
n, parts = *nn, nparts
}
}
// Internal iter function to walk nodes in lexicographical order.
func (t *SubjectTree[T]) iter(n node, pre []byte, ordered bool, cb func(subject []byte, val *T) bool) bool {
if n.isLeaf() {
ln := n.(*leaf[T])
return cb(append(pre, ln.suffix...), &ln.value)
}
// We are normal node here.
bn := n.base()
// Note that this append may reallocate, but it doesn't modify "pre" at the "iter" callsite.
pre = append(pre, bn.prefix...)
// Not everything requires lexicographical sorting, so support a fast path for iterating in
// whatever order the stree has things stored instead.
if !ordered {
for _, cn := range n.children() {
if cn == nil {
continue
}
if !t.iter(cn, pre, false, cb) {
return false
}
}
return true
}
// Collect nodes since unsorted.
var _nodes [256]node
nodes := _nodes[:0]
for _, cn := range n.children() {
if cn != nil {
nodes = append(nodes, cn)
}
}
// Now sort.
slices.SortStableFunc(nodes, func(a, b node) int { return bytes.Compare(a.path(), b.path()) })
// Now walk the nodes in order and call into next iter.
for i := range nodes {
if !t.iter(nodes[i], pre, true, cb) {
return false
}
}
return true
}
// LazyIntersect iterates the smaller of the two provided subject trees and
// looks for matching entries in the other. It is lazy in that it does not
// aggressively optimize against repeated walks, but is considerably faster
// in most cases than intersecting against a potentially large sublist.
func LazyIntersect[TL, TR any](tl *SubjectTree[TL], tr *SubjectTree[TR], cb func([]byte, *TL, *TR)) {
if tl == nil || tr == nil || tl.root == nil || tr.root == nil {
return
}
// Iterate over the smaller tree to reduce the number of rounds.
if tl.Size() <= tr.Size() {
tl.IterFast(func(key []byte, v1 *TL) bool {
if v2, ok := tr.Find(key); ok {
cb(key, v1, v2)
}
return true
})
} else {
tr.IterFast(func(key []byte, v2 *TR) bool {
if v1, ok := tl.Find(key); ok {
cb(key, v1, v2)
}
return true
})
}
}
// IntersectGSL will match all items in the given subject tree that
// have interest expressed in the given sublist. The callback will only be called
// once for each subject, regardless of overlapping subscriptions in the sublist.
func IntersectGSL[T any, SL comparable](t *SubjectTree[T], sl *gsl.GenericSublist[SL], cb func(subject []byte, val *T)) {
if t == nil || t.root == nil || sl == nil {
return
}
var _pre [256]byte
_intersectGSL(t.root, _pre[:0], sl, cb)
}
func _intersectGSL[T any, SL comparable](n node, pre []byte, sl *gsl.GenericSublist[SL], cb func(subject []byte, val *T)) {
if n.isLeaf() {
ln := n.(*leaf[T])
subj := append(pre, ln.suffix...)
if sl.HasInterest(bytesToString(subj)) {
cb(subj, &ln.value)
}
return
}
bn := n.base()
pre = append(pre, bn.prefix...)
for _, cn := range n.children() {
if cn == nil {
continue
}
subj := append(pre, cn.path()...)
if !hasInterestForTokens(sl, subj, len(pre)) {
continue
}
_intersectGSL(cn, pre, sl, cb)
}
}
// The subject tree can return partial tokens so we need to check starting interest
// only from whole tokens when we encounter a tsep.
func hasInterestForTokens[SL comparable](sl *gsl.GenericSublist[SL], subj []byte, since int) bool {
for i := since; i < len(subj); i++ {
if subj[i] == tsep {
if !sl.HasInterestStartingIn(bytesToString(subj[:i])) {
return false
}
}
}
return true
}
// Note this will avoid a copy of the data used for the string, but it will also reference the existing slice's data pointer.
// So this should be used sparingly when we know the encompassing byte slice's lifetime is the same.
func bytesToString(b []byte) string {
if len(b) == 0 {
return ""
}
p := unsafe.SliceData(b)
return unsafe.String(p, len(b))
}