/* ****************************************************************************** * Copyright (C) 2007, International Business Machines Corporation and * * others. All Rights Reserved. * ****************************************************************************** */ package com.ibm.icu.impl.duration.impl; import java.util.Locale; public class Utils { public static final Locale localeFromString(String s) { String language = s; String region = ""; String variant = ""; int x = language.indexOf("_"); if (x != -1) { region = language.substring(x+1); language = language.substring(0, x); } x = region.indexOf("_"); if (x != -1) { variant = region.substring(x+1); region = region.substring(0, x); } return new Locale(language, region, variant); } /* public static T[] arraycopy(T[] src) { T[] result = (T[])Array.newInstance(src.getClass().getComponentType(), src.length); // can we do this without casting? for (int i = 0; i < src.length; ++i) { result[i] = src[i]; } return result; } */ /** * Interesting features of chinese numbers: * - Each digit is followed by a unit symbol (10's, 100's, 1000's). * - Units repeat in levels of 10,000, there are symbols for each level too (except 1's). * - The digit 2 has a special form before the 10 symbol and at the end of the number. * - If the first digit in the number is 1 and its unit is 10, the 1 is omitted. * - Sequences of 0 digits and their units are replaced by a single 0 and no unit. * - If there are two such sequences of 0 digits in a level (1000's and 10's), the 1000's 0 is also omitted. * - The 1000's 0 is also omitted in alternating levels, such that it is omitted in the rightmost * level with a 10's 0, or if none, in the rightmost level. * - Level symbols are omitted if all of their units are omitted */ public static String chineseNumber(long n, ChineseDigits zh) { if (n < 0) { n = -n; } if (n <= 10) { if (n == 2) { return String.valueOf(zh.liang); } return String.valueOf(zh.digits[(int)n]); } // 9223372036854775807 char[] buf = new char[40]; // as long as we get, and actually we can't get this high, no units past zhao char[] digits = String.valueOf(n).toCharArray(); // first, generate all the digits in place // convert runs of zeros into a single zero, but keep places // boolean inZero = true; // true if we should zap zeros in this block, resets at start of block boolean forcedZero = false; // true if we have a 0 in tens's place int x = buf.length; for (int i = digits.length, u = -1, l = -1; --i >= 0;) { if (u == -1) { if (l != -1) { buf[--x] = zh.levels[l]; inZero = true; forcedZero = false; } ++u; } else { buf[--x] = zh.units[u++]; if (u == 3) { u = -1; ++l; } } int d = digits[i] - '0'; if (d == 0) { if (x < buf.length-1 && u != 0) { buf[x] = '*'; } if (inZero || forcedZero) { buf[--x] = '*'; } else { buf[--x] = zh.digits[0]; inZero = true; forcedZero = u == 1; } } else { inZero = false; buf[--x] = zh.digits[d]; } } // scanning from right, find first required 'ling' // we only care if n > 101,0000 as this is the first case where // it might shift. remove optional lings in alternating blocks. if (n > 1000000) { boolean last = true; int i = buf.length - 3; do { if (buf[i] == '0') { break; } i -= 8; last = !last; } while (i > x); i = buf.length - 7; do { if (buf[i] == zh.digits[0] && !last) { buf[i] = '*'; } i -= 8; last = !last; } while (i > x); // remove levels for empty blocks if (n >= 100000000) { i = buf.length - 8; do { boolean empty = true; for (int j = i-1, e = Math.max(x-1, i-8); j > e; --j) { if (buf[j] != '*') { empty = false; break; } } if (empty) { if (buf[i+1] != '*' && buf[i+1] != zh.digits[0]) { buf[i] = zh.digits[0]; } else { buf[i] = '*'; } } i -= 8; } while (i > x); } } // replace er by liang except before or after shi or after ling for (int i = x; i < buf.length; ++i) { if (buf[i] != zh.digits[2]) continue; if (i < buf.length - 1 && buf[i+1] == zh.units[0]) continue; if (i > x && (buf[i-1] == zh.units[0] || buf[i-1] == zh.digits[0] || buf[i-1] == '*')) continue; buf[i] = zh.liang; } // eliminate leading 1 if following unit is shi if (buf[x] == zh.digits[1] && (zh.ko || buf[x+1] == zh.units[0])) { ++x; } // now, compress out the '*' int w = x; for (int r = x; r < buf.length; ++r) { if (buf[r] != '*') { buf[w++] = buf[r]; } } return new String(buf, x, w-x); } public static void main(String[] args) { for (int i = 0; i < args.length; ++i) { String arg = args[i]; System.out.print(arg); System.out.print(" > "); long n = Long.parseLong(arg); System.out.println(chineseNumber(n, ChineseDigits.DEBUG)); } } public static class ChineseDigits { final char[] digits; final char[] units; final char[] levels; final char liang; final boolean ko; ChineseDigits(String digits, String units, String levels, char liang, boolean ko) { this.digits = digits.toCharArray(); this.units = units.toCharArray(); this.levels = levels.toCharArray(); this.liang = liang; this.ko = ko; } public static final ChineseDigits DEBUG = new ChineseDigits("0123456789s", "sbq", "WYZ", 'L', false); public static final ChineseDigits TRADITIONAL = new ChineseDigits("\u96f6\u4e00\u4e8c\u4e09\u56db\u4e94\u516d\u4e03\u516b\u4e5d\u5341", // to shi "\u5341\u767e\u5343", // shi, bai, qian "\u842c\u5104\u5146", // wan, yi, zhao '\u5169', false); // liang public static final ChineseDigits SIMPLIFIED = new ChineseDigits("\u96f6\u4e00\u4e8c\u4e09\u56db\u4e94\u516d\u4e03\u516b\u4e5d\u5341", // to shi "\u5341\u767e\u5343", // shi, bai, qian "\u4e07\u4ebf\u5146", // wan, yi, zhao '\u4e24', false); // liang // no 1 before first unit no matter what it is // not sure if there are 'ling' units public static final ChineseDigits KOREAN = new ChineseDigits("\uc601\uc77c\uc774\uc0bc\uc0ac\uc624\uc721\uce60\ud314\uad6c\uc2ed", // to ten "\uc2ed\ubc31\ucc9c", // 10, 100, 1000 "\ub9cc\uc5b5?", // 10^4, 10^8, 10^12 '\uc774', true); } }