我的CharsetDecoder类有问题。

代码的第一个示例(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
但是当我执行此代码时:
    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?

如何使用类decode(ByteBuffer, CharBuffer, endOfInput)CharsetDecoder方法来检索结果a€a

-编辑-

因此,使用Jesper的代码可以做到这一点。这不是完美的方法,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());

最佳答案

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

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

您将看到以下输出:
UNDERFLOW
MALFORMED[1]
MALFORMED[1]
a   a

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

编辑-解码器报告UNDERFLOW时,它希望您将更多数据添加到输入缓冲区,然后尝试再次调用decode(),但是必须从要尝试的UTF-8序列的开头重新提供数据。解码。您不能在UTF-8序列的中间继续。

这是一个有效的版本,在循环的每次迭代中从tab添加一个字节:
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 €

final ByteBuffer buffer = ByteBuffer.allocate(10);

for (int i = 0; i < tab.length; i++) {
    // Add the next byte to the buffer
    buffer.put(tab[i]);

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

    // Try to decode
    buffer.flip();
    final CoderResult result = dec.decode(buffer, chars, i == 2);
    System.out.println(result);

    if (result.isUnderflow()) {
        // Underflow, prepare the buffer for more writing
        buffer.limit(buffer.capacity());
        buffer.position(pos);
    }
}

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

System.out.println("a" + chars.toString() + "a");

关于java - 什么是CharsetDecoder.decode(ByteBuffer,CharBuffer,endOfInput),我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/29559842/

10-14 04:28