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package iradix import ( "bytes" ) // Iterator is used to iterate over a set of nodes // in pre-order type Iterator struct { node *Node stack []edges } // SeekPrefixWatch is used to seek the iterator to a given prefix // and returns the watch channel of the finest granularity func (i *Iterator) SeekPrefixWatch(prefix []byte) (watch <-chan struct{}) { // Wipe the stack i.stack = nil n := i.node watch = n.mutateCh search := prefix for { // Check for key exhaustion if len(search) == 0 { i.node = n return } // Look for an edge _, n = n.getEdge(search[0]) if n == nil { i.node = nil return } // Update to the finest granularity as the search makes progress watch = n.mutateCh // Consume the search prefix if bytes.HasPrefix(search, n.prefix) { search = search[len(n.prefix):] } else if bytes.HasPrefix(n.prefix, search) { i.node = n return } else { i.node = nil return } } } // SeekPrefix is used to seek the iterator to a given prefix func (i *Iterator) SeekPrefix(prefix []byte) { i.SeekPrefixWatch(prefix) } func (i *Iterator) recurseMin(n *Node) *Node { // Traverse to the minimum child if n.leaf != nil { return n } nEdges := len(n.edges) if nEdges > 1 { // Add all the other edges to the stack (the min node will be added as // we recurse) i.stack = append(i.stack, n.edges[1:]) } if nEdges > 0 { return i.recurseMin(n.edges[0].node) } // Shouldn't be possible return nil } // SeekLowerBound is used to seek the iterator to the smallest key that is // greater or equal to the given key. There is no watch variant as it's hard to // predict based on the radix structure which node(s) changes might affect the // result. func (i *Iterator) SeekLowerBound(key []byte) { // Wipe the stack. Unlike Prefix iteration, we need to build the stack as we // go because we need only a subset of edges of many nodes in the path to the // leaf with the lower bound. Note that the iterator will still recurse into // children that we don't traverse on the way to the reverse lower bound as it // walks the stack. i.stack = []edges{} // i.node starts off in the common case as pointing to the root node of the // tree. By the time we return we have either found a lower bound and setup // the stack to traverse all larger keys, or we have not and the stack and // node should both be nil to prevent the iterator from assuming it is just // iterating the whole tree from the root node. Either way this needs to end // up as nil so just set it here. n := i.node i.node = nil search := key found := func(n *Node) { i.stack = append(i.stack, edges{edge{node: n}}) } findMin := func(n *Node) { n = i.recurseMin(n) if n != nil { found(n) return } } for { // Compare current prefix with the search key's same-length prefix. var prefixCmp int if len(n.prefix) < len(search) { prefixCmp = bytes.Compare(n.prefix, search[0:len(n.prefix)]) } else { prefixCmp = bytes.Compare(n.prefix, search) } if prefixCmp > 0 { // Prefix is larger, that means the lower bound is greater than the search // and from now on we need to follow the minimum path to the smallest // leaf under this subtree. findMin(n) return } if prefixCmp < 0 { // Prefix is smaller than search prefix, that means there is no lower // bound i.node = nil return } // Prefix is equal, we are still heading for an exact match. If this is a // leaf and an exact match we're done. if n.leaf != nil && bytes.Equal(n.leaf.key, key) { found(n) return } // Consume the search prefix if the current node has one. Note that this is // safe because if n.prefix is longer than the search slice prefixCmp would // have been > 0 above and the method would have already returned. search = search[len(n.prefix):] if len(search) == 0 { // We've exhausted the search key, but the current node is not an exact // match or not a leaf. That means that the leaf value if it exists, and // all child nodes must be strictly greater, the smallest key in this // subtree must be the lower bound. findMin(n) return } // Otherwise, take the lower bound next edge. idx, lbNode := n.getLowerBoundEdge(search[0]) if lbNode == nil { return } // Create stack edges for the all strictly higher edges in this node. if idx+1 < len(n.edges) { i.stack = append(i.stack, n.edges[idx+1:]) } // Recurse n = lbNode } } // Next returns the next node in order func (i *Iterator) Next() ([]byte, interface{}, bool) { // Initialize our stack if needed if i.stack == nil && i.node != nil { i.stack = []edges{ { edge{node: i.node}, }, } } for len(i.stack) > 0 { // Inspect the last element of the stack n := len(i.stack) last := i.stack[n-1] elem := last[0].node // Update the stack if len(last) > 1 { i.stack[n-1] = last[1:] } else { i.stack = i.stack[:n-1] } // Push the edges onto the frontier if len(elem.edges) > 0 { i.stack = append(i.stack, elem.edges) } // Return the leaf values if any if elem.leaf != nil { return elem.leaf.key, elem.leaf.val, true } } return nil, nil, false }