AST Tweaks (#551)
* Use pointers instead of copying around ast.Node Node is a 56B struct that is constantly in the hot path. Passing nodes around by copy had a cost that started to add up. This change replaces them by pointers. Using unsafe pointer arithmetic and converting sibling/child indexes to relative offsets, it removes the need to carry around a pointer to the root of the tree. This saves 8B per Node. This space will be used to store an extra []byte slice to provide contextual error handling on all nodes, including the ones whose data is different than the raw input (for example: strings with escaped characters), while staying under the size of a cache line. * Remove conditional * Add Raw to track range in data for parsed values * Simplify reference tracking
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+11
-20
@@ -1,12 +1,11 @@
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package ast
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type Reference struct {
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idx int
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set bool
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}
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type Reference int
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const InvalidReference Reference = -1
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func (r Reference) Valid() bool {
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return r.set
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return r != InvalidReference
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}
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type Builder struct {
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@@ -18,8 +17,8 @@ func (b *Builder) Tree() *Root {
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return &b.tree
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}
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func (b *Builder) NodeAt(ref Reference) Node {
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return b.tree.at(ref.idx)
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func (b *Builder) NodeAt(ref Reference) *Node {
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return b.tree.at(ref)
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}
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func (b *Builder) Reset() {
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@@ -28,33 +27,25 @@ func (b *Builder) Reset() {
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}
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func (b *Builder) Push(n Node) Reference {
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n.root = &b.tree
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b.lastIdx = len(b.tree.nodes)
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b.tree.nodes = append(b.tree.nodes, n)
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return Reference{
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idx: b.lastIdx,
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set: true,
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}
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return Reference(b.lastIdx)
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}
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func (b *Builder) PushAndChain(n Node) Reference {
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n.root = &b.tree
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newIdx := len(b.tree.nodes)
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b.tree.nodes = append(b.tree.nodes, n)
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if b.lastIdx >= 0 {
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b.tree.nodes[b.lastIdx].next = newIdx
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b.tree.nodes[b.lastIdx].next = newIdx - b.lastIdx
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}
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b.lastIdx = newIdx
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return Reference{
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idx: b.lastIdx,
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set: true,
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}
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return Reference(b.lastIdx)
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}
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func (b *Builder) AttachChild(parent Reference, child Reference) {
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b.tree.nodes[parent.idx].child = child.idx
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b.tree.nodes[parent].child = int(child) - int(parent)
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}
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func (b *Builder) Chain(from Reference, to Reference) {
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b.tree.nodes[from.idx].next = to.idx
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b.tree.nodes[from].next = int(to) - int(from)
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}
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