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Java example source code file (FindASCIIRangeCodingBugs.java)
The FindASCIIRangeCodingBugs.java Java example source code/* * Copyright (c) 2008, Oracle and/or its affiliates. All rights reserved. * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER. * * This code is free software; you can redistribute it and/or modify it * under the terms of the GNU General Public License version 2 only, as * published by the Free Software Foundation. * * This code is distributed in the hope that it will be useful, but WITHOUT * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License * version 2 for more details (a copy is included in the LICENSE file that * accompanied this code). * * You should have received a copy of the GNU General Public License version * 2 along with this work; if not, write to the Free Software Foundation, * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA. * * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA * or visit www.oracle.com if you need additional information or have any * questions. */ /* @test * @bug 6378295 * @summary Roundtrip Encoding/Decoding of ASCII chars from 0x00-0x7f */ import java.util.*; import java.nio.*; import java.nio.charset.*; public class FindASCIIRangeCodingBugs { private static int failures = 0; private static byte[] asciiBytes = new byte[0x80]; private static char[] asciiChars = new char[0x80]; private static String asciiString; private static void check(String csn) throws Exception { System.out.println(csn); if (! Arrays.equals(asciiString.getBytes(csn), asciiBytes)) { System.out.printf("%s -> bytes%n", csn); failures++; } if (! new String(asciiBytes, csn).equals(asciiString)) { System.out.printf("%s -> chars%n", csn); failures++; } } public static void main(String[] args) throws Exception { for (int i = 0; i < 0x80; i++) { asciiBytes[i] = (byte) i; asciiChars[i] = (char) i; } asciiString = new String(asciiChars); Charset ascii = Charset.forName("ASCII"); for (Map.Entry<String,Charset> e : Charset.availableCharsets().entrySet()) { String csn = e.getKey(); Charset cs = e.getValue(); if (!cs.contains(ascii) || csn.matches(".*2022.*") || //iso2022 family csn.matches("x-windows-5022[0|1]") || //windows 2022jp csn.matches(".*UTF-[16|32].*")) //multi-bytes continue; if (! cs.canEncode()) continue; try { check(csn); } catch (Throwable t) { t.printStackTrace(); failures++; } } if (failures > 0) throw new Exception(failures + "tests failed"); } } Other Java examples (source code examples)Here is a short list of links related to this Java FindASCIIRangeCodingBugs.java source code file: |
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