Java字节缓冲区getLong()溢出long变量

Question

我正在使用非常简单的骨头ByteBuffer与java 1.4一起使用.我有一个小骨架,基本上只有糟糕的put/getInt()put/getLong()实现.我的问题是,当putInt和getInt工作时,getLong()(我认为它)不起作用.

当我读出第四个字节并将其移入长时间溢出时.但是我所有的变量都很长,所以它不应该溢出.

下面是代码(请记住,这只是一个开始):

public class ByteBuffer {

    private byte[] buffer;
    private int first = 0;
    private int last = 0;
    private int size;
    private int elements;

    public ByteBuffer(int size) {
        this.size = size;
        buffer = new byte[size];
    }

    public void putInt(int theInt) {
        for (int i = 0; i < 4; i++) {
            put((byte) (theInt >>> (8 * i)));
        }
    }

    public int getInt() {
        int theInt = 0;
        for (int i = 0; i < 4; i++) {
            theInt |= (0xFF & get()) << (8 * i);
        }
        return theInt;
    }

    public void putLong(long theLong) {
        for (int i = 0; i < 8; i++) {
            put((byte) (theLong >>> (8 * i)));
        }
    }

    public long getLong() {
        long theLong = 0L;
        for (int i = 0; i < 8; i++) {
            theLong |= (long) ((0xFF & get()) << (8 * i));
        }
        return theLong;
    }

    public void put(byte theByte) {
        buffer[last++] = theByte;
        if (last == size) {
            last = 0;
        }
        elements++;
    }

    public byte get() {
        byte theByte = buffer[first++];
        if (first == size) {
            first = 0;
        }
        elements--;
        return theByte;
    }

    public byte[] array() {
        return (byte[]) buffer.clone();
    }

    /**
     * @param args
     */
    public static void main(String[] args) {
        ByteBuffer buff = new ByteBuffer(512);

        buff.putLong(9223372036854775807L);
        buff.putLong(-9223372036854775808L);

        System.out.println(9223372036854775807L);
        System.out.println(-9223372036854775808L);

        long l1 = buff.getLong();
        long l2 = buff.getLong();
        System.out.println(l1);
        System.out.println(l2);
    }

}

Answer 1

在你的getLong方法中,你必须将(0xFF和get())转换为long,然后才能将它移动超过32位.您还可以使用长文字(0xFFL)而不是int literal(0xFF).

这样做的原因是"int && byte"操作的结果类型(0xFF&get())是一个int.位移操作的规范是这样的,如果a是int,则"a << b"实际上将移位"b模32"位,如果a是long,则"b模64"位.