Cursor Agent
f09f77ab06
Remove internal/danger package and replace unsafe pointer arithmetic with direct pointer manipulation. Update AST node references to use pointers instead of integer offsets. This improves code safety and maintainability. Co-authored-by: thomas.pelletier <thomas.pelletier@bedrockrobotics.com>
92 lines
1.9 KiB
Go
92 lines
1.9 KiB
Go
package unstable
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// root contains a full AST.
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//
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// It is immutable once constructed with Builder.
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type root struct {
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nodes []Node
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}
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// Iterator over the top level nodes.
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func (r *root) Iterator() Iterator {
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it := Iterator{}
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if len(r.nodes) > 0 {
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it.node = &r.nodes[0]
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}
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return it
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}
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func (r *root) at(idx reference) *Node {
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return &r.nodes[idx]
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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 != invalidReference
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}
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type builder struct {
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tree root
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lastIdx int
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nextOffsets []int
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childOffsets []int
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}
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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)
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}
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func (b *builder) Reset() {
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b.tree.nodes = b.tree.nodes[:0]
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b.nextOffsets = b.nextOffsets[:0]
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b.childOffsets = b.childOffsets[:0]
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b.lastIdx = 0
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}
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func (b *builder) Push(n Node) reference {
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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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b.nextOffsets = append(b.nextOffsets, 0)
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b.childOffsets = append(b.childOffsets, 0)
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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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newIdx := len(b.tree.nodes)
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b.tree.nodes = append(b.tree.nodes, n)
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b.nextOffsets = append(b.nextOffsets, 0)
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b.childOffsets = append(b.childOffsets, 0)
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if b.lastIdx >= 0 {
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b.nextOffsets[b.lastIdx] = newIdx - b.lastIdx
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}
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b.lastIdx = newIdx
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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.childOffsets[parent] = 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.nextOffsets[from] = int(to) - int(from)
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}
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func (b *builder) Link() {
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for i := range b.tree.nodes {
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if next := b.nextOffsets[i]; next != 0 {
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b.tree.nodes[i].next = &b.tree.nodes[i+next]
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}
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if child := b.childOffsets[i]; child != 0 {
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b.tree.nodes[i].child = &b.tree.nodes[i+child]
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}
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}
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}
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