什么是CharsetDecoder.decode(ByteBuffer,CharBuffer,endOfInput)

Question

我CharsetDecoder上课有问题.

代码的第一个例子(有效):

    final CharsetDecoder dec = Charset.forName("UTF-8").newDecoder();
    final ByteBuffer b = ByteBuffer.allocate(3);
    final byte[] tab = new byte[]{(byte)-30, (byte)-126, (byte)-84}; //char €
    for (int i=0; i<tab.length; i++){
        b.put(tab, i, 1);
    }
    try {
        b.flip();
        System.out.println("a" + dec.decode(b).toString() + "a");
    } catch (CharacterCodingException e1) {
        e1.printStackTrace();
    }

结果是 a€a

但是当我执行这段代码时:

    final CharsetDecoder dec = Charset.forName("UTF-8").newDecoder();
    final CharBuffer chars = CharBuffer.allocate(3);
    final byte[] tab = new byte[]{(byte)-30, (byte)-126, (byte)-84}; //char €
    for (int i=0; i<tab.length; i++){
        ByteBuffer buffer = ByteBuffer.wrap(tab, i, 1);
        dec.decode(buffer, chars, i == 2);
    }
    dec.flush(chars);
    System.out.println("a" + chars.toString() + "a");

结果是 a

为什么结果不一样？

如何使用decode(ByteBuffer, CharBuffer, endOfInput)类的方法CharsetDecoder来检索结果a€a？

- 编辑 -

所以使用Jesper的代码我这样做.这不是完美的,但可以使用step= 1,2和3

final CharsetDecoder dec = Charset.forName("UTF-8").newDecoder();
    final CharBuffer chars = CharBuffer.allocate(6);
    final byte[] tab = new byte[]{(byte)97, (byte)-30, (byte)-126, (byte)-84, (byte)97, (byte)97}; //char €

    final ByteBuffer buffer = ByteBuffer.allocate(10);

    final int step = 3;
    for (int i = 0; i < tab.length; i++) {
        // Add the next byte to the buffer
        buffer.put(tab, i, step);
        i+=step-1;

        // Remember the current position
        final int pos = buffer.position();
        int l=chars.position();

        // Try to decode
        buffer.flip();
        final CoderResult result = dec.decode(buffer, chars, i >= tab.length -1);
        System.out.println(result);

        if (result.isUnderflow() && chars.position() == l) {
            // Underflow, prepare the buffer for more writing
            buffer.position(pos);
        }else{
            if (buffer.position() == buffer.limit()){
                //ByteBuffer decoded
                buffer.clear();
                buffer.position(0);
            }else{
                //a part of ByteBuffer is decoded. We keep only bytes which are not decoded
                final byte[] b = buffer.array();
                final int f = buffer.position();
                final int g = buffer.limit() - buffer.position();
                buffer.clear();
                buffer.position(0);
                buffer.put(b, f, g);
            }
        }
        buffer.limit(buffer.capacity());
    }

    dec.flush(chars);
    chars.flip();

    System.out.println(chars.toString());

Answer 1

该方法decode(ByteBuffer, CharBuffer, boolean)返回一个结果，但您忽略了该结果。如果在第二个代码片段中打印结果：

\n\n

for (int i = 0; i < tab.length; i++) {\n    ByteBuffer buffer = ByteBuffer.wrap(tab, i, 1);\n    System.out.println(dec.decode(buffer, chars, i == 2));\n}\n

\n\n

你会看到这个输出：

\n\n

UNDERFLOW\nMALFORMED[1]\nMALFORMED[1]\na   a\n

\n\n

显然，如果您在字符中间开始解码，它就无法正常工作。解码器期望它读取的第一件事是有效 UTF-8 序列的开头。

\n\n

编辑- 当解码器报告 时UNDERFLOW，它希望您向输入缓冲区添加更多数据，然后尝试再次调用decode()，但您必须重新向其提供您尝试解码的 UTF-8 序列开头的数据。您无法在 UTF-8 序列的中间继续。

\n\n

这是一个有效的版本，tab在循环的每次迭代中添加一个字节：

\n\n

final CharsetDecoder dec = Charset.forName("UTF-8").newDecoder();\nfinal CharBuffer chars = CharBuffer.allocate(3);\nfinal byte[] tab = new byte[]{(byte) -30, (byte) -126, (byte) -84}; //char \xe2\x82\xac\n\nfinal ByteBuffer buffer = ByteBuffer.allocate(10);\n\nfor (int i = 0; i < tab.length; i++) {\n    // Add the next byte to the buffer\n    buffer.put(tab[i]);\n\n    // Remember the current position\n    final int pos = buffer.position();\n\n    // Try to decode\n    buffer.flip();\n    final CoderResult result = dec.decode(buffer, chars, i == 2);\n    System.out.println(result);\n\n    if (result.isUnderflow()) {\n        // Underflow, prepare the buffer for more writing\n        buffer.limit(buffer.capacity());\n        buffer.position(pos);\n    }\n}\n\ndec.flush(chars);\nchars.flip();\n\nSystem.out.println("a" + chars.toString() + "a");\n

\n