--- /dev/null
+/*
+ *******************************************************************************
+ * Copyright (C) 2013, International Business Machines Corporation and *
+ * others. All Rights Reserved. *
+ *******************************************************************************
+ */
+package com.ibm.icu.text;
+
+import java.io.IOException;
+import java.text.CharacterIterator;
+import java.util.Stack;
+
+import com.ibm.icu.lang.UCharacter;
+import com.ibm.icu.lang.UProperty;
+import com.ibm.icu.lang.UScript;
+
+class LaoBreakEngine implements LanguageBreakEngine {
+ /* Helper class for improving readability of the Lao word break
+ * algorithm.
+ */
+ static class PossibleWord {
+ // List size, limited by the maximum number of words in the dictionary
+ // that form a nested sequence.
+ private final static int POSSIBLE_WORD_LIST_MAX = 20;
+ //list of word candidate lengths, in increasing length order
+ private int lengths[];
+ private int count[]; // Count of candidates
+ private int prefix; // The longest match with a dictionary word
+ private int offset; // Offset in the text of these candidates
+ private int mark; // The preferred candidate's offset
+ private int current; // The candidate we're currently looking at
+
+ // Default constructor
+ public PossibleWord() {
+ lengths = new int[POSSIBLE_WORD_LIST_MAX];
+ count = new int[1]; // count needs to be an array of 1 so that it can be pass as reference
+ offset = -1;
+ }
+
+ // Fill the list of candidates if needed, select the longest, and return the number found
+ public int candidates(CharacterIterator fIter, DictionaryMatcher dict, int rangeEnd) {
+ int start = fIter.getIndex();
+ if (start != offset) {
+ offset = start;
+ prefix = dict.matches(fIter, rangeEnd - start, lengths, count, lengths.length);
+ // Dictionary leaves text after longest prefix, not longest word. Back up.
+ if (count[0] <= 0) {
+ fIter.setIndex(start);
+ }
+ }
+ if (count[0] > 0) {
+ fIter.setIndex(start + lengths[count[0]-1]);
+ }
+ current = count[0] - 1;
+ mark = current;
+ return count[0];
+ }
+
+ // Select the currently marked candidate, point after it in the text, and invalidate self
+ public int acceptMarked(CharacterIterator fIter) {
+ fIter.setIndex(offset + lengths[mark]);
+ return lengths[mark];
+ }
+
+ // Backup from the current candidate to the next shorter one; return true if that exists
+ // and point the text after it
+ public boolean backUp(CharacterIterator fIter) {
+ if (current > 0) {
+ fIter.setIndex(offset + lengths[--current]);
+ return true;
+ }
+ return false;
+ }
+
+ // Return the longest prefix this candidate location shares with a dictionary word
+ public int longestPrefix() {
+ return prefix;
+ }
+
+ // Mark the current candidate as the one we like
+ public void markCurrent() {
+ mark = current;
+ }
+ }
+
+ // Constants for LaoBreakIterator
+ // How many words in a row are "good enough"?
+ private static final byte LAO_LOOKAHEAD = 3;
+ // Will not combine a non-word with a preceding dictionary word longer than this
+ private static final byte LAO_ROOT_COMBINE_THRESHOLD = 3;
+ // Will not combine a non-word that shares at least this much prefix with a
+ // dictionary word with a preceding word
+ private static final byte LAO_PREFIX_COMBINE_THRESHOLD = 3;
+ // Minimum word size
+ private static final byte LAO_MIN_WORD = 2;
+
+ private DictionaryMatcher fDictionary;
+ private static UnicodeSet fLaoWordSet;
+ private static UnicodeSet fEndWordSet;
+ private static UnicodeSet fBeginWordSet;
+ private static UnicodeSet fMarkSet;
+
+ static {
+ // Initialize UnicodeSets
+ fLaoWordSet = new UnicodeSet();
+ fMarkSet = new UnicodeSet();
+ fEndWordSet = new UnicodeSet();
+ fBeginWordSet = new UnicodeSet();
+
+ fLaoWordSet.applyPattern("[[:Laoo:]&[:LineBreak=SA:]]");
+ fLaoWordSet.compact();
+
+ fMarkSet.applyPattern("[[:Laoo:]&[:LineBreak=SA:]&[:M:]]");
+ fMarkSet.add(0x0020);
+ fEndWordSet = fLaoWordSet;
+ fEndWordSet.remove(0x0EC0, 0x0EC4); // prefix vowels
+ fBeginWordSet.add(0x0E81, 0x0EAE); // basic consonants (including holes for corresponding Thai characters)
+ fBeginWordSet.add(0x0EDC, 0x0EDD); // digraph consonants (no Thai equivalent)
+ fBeginWordSet.add(0x0EC0, 0x0EC4); // prefix vowels
+
+ // Compact for caching
+ fMarkSet.compact();
+ fEndWordSet.compact();
+ fBeginWordSet.compact();
+
+ // Freeze the static UnicodeSet
+ fLaoWordSet.freeze();
+ fMarkSet.freeze();
+ fEndWordSet.freeze();
+ fBeginWordSet.freeze();
+ }
+
+ public LaoBreakEngine() throws IOException {
+ // Initialize dictionary
+ fDictionary = DictionaryData.loadDictionaryFor("Laoo");
+ }
+
+ public boolean handles(int c, int breakType) {
+ if (breakType == BreakIterator.KIND_WORD || breakType == BreakIterator.KIND_LINE) {
+ int script = UCharacter.getIntPropertyValue(c, UProperty.SCRIPT);
+ return (script == UScript.LAO);
+ }
+ return false;
+ }
+
+ public int findBreaks(CharacterIterator fIter, int rangeStart, int rangeEnd, boolean reverse, int breakType,
+ Stack<Integer> foundBreaks) {
+ if ((rangeEnd - rangeStart) < LAO_MIN_WORD) {
+ return 0; // Not enough characters for word
+ }
+ int wordsFound = 0;
+ int wordLength;
+ int current;
+ PossibleWord words[] = new PossibleWord[LAO_LOOKAHEAD];
+ for (int i = 0; i < LAO_LOOKAHEAD; i++) {
+ words[i] = new PossibleWord();
+ }
+ int uc;
+
+ fIter.setIndex(rangeStart);
+
+ while ((current = fIter.getIndex()) < rangeEnd) {
+ wordLength = 0;
+
+ //Look for candidate words at the current position
+ int candidates = words[wordsFound%LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd);
+
+ // If we found exactly one, use that
+ if (candidates == 1) {
+ wordLength = words[wordsFound%LAO_LOOKAHEAD].acceptMarked(fIter);
+ wordsFound += 1;
+ }
+
+ // If there was more than one, see which one can take us forward the most words
+ else if (candidates > 1) {
+ boolean foundBest = false;
+ // If we're already at the end of the range, we're done
+ if (fIter.getIndex() < rangeEnd) {
+ do {
+ int wordsMatched = 1;
+ if (words[(wordsFound+1)%LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd) > 0) {
+ if (wordsMatched < 2) {
+ // Followed by another dictionary word; mark first word as a good candidate
+ words[wordsFound%LAO_LOOKAHEAD].markCurrent();
+ wordsMatched = 2;
+ }
+
+ // If we're already at the end of the range, we're done
+ if (fIter.getIndex() >= rangeEnd) {
+ break;
+ }
+
+ // See if any of the possible second words is followed by a third word
+ do {
+ // If we find a third word, stop right away
+ if (words[(wordsFound+2)%LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd) > 0) {
+ words[wordsFound%LAO_LOOKAHEAD].markCurrent();
+ foundBest = true;
+ break;
+ }
+ } while (words[(wordsFound+1)%LAO_LOOKAHEAD].backUp(fIter));
+ }
+ } while (words[wordsFound%LAO_LOOKAHEAD].backUp(fIter) && !foundBest);
+ }
+ wordLength = words[wordsFound%LAO_LOOKAHEAD].acceptMarked(fIter);
+ wordsFound += 1;
+ }
+
+ // We come here after having either found a word or not. We look ahead to the
+ // next word. If it's not a dictionary word, we will combine it with the word we
+ // just found (if there is one), but only if the preceding word does not exceed
+ // the threshold.
+ // The text iterator should now be positioned at the end of the word we found.
+ if (fIter.getIndex() < rangeEnd && wordLength < LAO_ROOT_COMBINE_THRESHOLD) {
+ // If it is a dictionary word, do nothing. If it isn't, then if there is
+ // no preceding word, or the non-word shares less than the minimum threshold
+ // of characters with a dictionary word, then scan to resynchronize
+ if (words[wordsFound%LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd) <= 0 &&
+ (wordLength == 0 ||
+ words[wordsFound%LAO_LOOKAHEAD].longestPrefix() < LAO_PREFIX_COMBINE_THRESHOLD)) {
+ // Look for a plausible word boundary
+ int remaining = rangeEnd - (current + wordLength);
+ int pc = fIter.current();
+ int chars = 0;
+ for (;;) {
+ fIter.next();
+ uc = fIter.current();
+ chars += 1;
+ if (--remaining <= 0) {
+ break;
+ }
+ if (fEndWordSet.contains(pc) && fBeginWordSet.contains(uc)) {
+ // Maybe. See if it's in the dictionary.
+ int candidate = words[(wordsFound + 1) %LAO_LOOKAHEAD].candidates(fIter, fDictionary, rangeEnd);
+ fIter.setIndex(current + wordLength + chars);
+ if (candidate > 0) {
+ break;
+ }
+ }
+ pc = uc;
+ }
+
+ // Bump the word count if there wasn't already one
+ if (wordLength <= 0) {
+ wordsFound += 1;
+ }
+
+ // Update the length with the passed-over characters
+ wordLength += chars;
+ } else {
+ // Backup to where we were for next iteration
+ fIter.setIndex(current+wordLength);
+ }
+ }
+
+ // Never stop before a combining mark.
+ int currPos;
+ while ((currPos = fIter.getIndex()) < rangeEnd && fMarkSet.contains(fIter.current())) {
+ fIter.next();
+ wordLength += fIter.getIndex() - currPos;
+ }
+
+ // Look ahead for possible suffixes if a dictionary word does not follow.
+ // We do this in code rather than using a rule so that the heuristic
+ // resynch continues to function. For example, one of the suffix characters
+ // could be a typo in the middle of a word.
+ // NOT CURRENTLY APPLICABLE TO LAO
+
+ // Did we find a word on this iteration? If so, push it on the break stack
+ if (wordLength > 0) {
+ foundBreaks.push(Integer.valueOf(current + wordLength));
+ }
+ }
+
+ // Don't return a break for the end of the dictionary range if there is one there
+ if (foundBreaks.peek().intValue() >= rangeEnd) {
+ foundBreaks.pop();
+ wordsFound -= 1;
+ }
+
+ return wordsFound;
+ }
+
+}
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