是否可以通过确定要读取的数量来循环读取位？

Question

有一个 DataInputStream 从中读取我遇到了一个特定的算法：int[] nmbrs = new int[64];读取一个 4 位无符号整数。将读取的值分配给长度对于 (int i = 0; i < 64; i++)3.1 读取一个长度位的无符号整数作为nmbrs[i]可以用Java写吗？怎么写呢？

Answer 1

可以用Java写吗？怎么写呢？

Java 没有提供以小于一个字节的单位执行 I/O 的机制，但您可以在面向字节的 I/O 之上实现它。您需要在读取时一次缓冲一个或多个字节，并跟踪该缓冲区内的位级位置。

另请注意，这对（逻辑）位顺序问题很敏感——即您是从最高位读到最低位还是相反？

Answer 2

创建一个BitInputStream从底层读取位的类DataInputStream。

像这样：

public final class BitInputStream implements Closeable {

    private final InputStream in;

    private final ByteOrder streamBitOrder;

    private int bits;

    private byte bitsLeft;

    public BitInputStream(InputStream in) {

        this(in, ByteOrder.BIG_ENDIAN);

    }

    public BitInputStream(InputStream in, ByteOrder bitOrder) {

        Objects.requireNonNull(in);

        Objects.requireNonNull(bitOrder);

        this.in = in;

        this.streamBitOrder = bitOrder;

    }

    @Override

    public void close() throws IOException {

        this.in.close();

    }

    public int readBit() throws IOException {

        if (this.bitsLeft == 0) {

            if ((this.bits = this.in.read()) == -1)

                throw new EOFException();

            this.bitsLeft = 8;

        }

        int bitIdx = (this.streamBitOrder == ByteOrder.BIG_ENDIAN ? this.bitsLeft - 1 : 8 - this.bitsLeft);

        this.bitsLeft--;

        return (this.bits >> bitIdx) & 1;

    }

    public int readInt() throws IOException {

        return readInt(Integer.SIZE, this.streamBitOrder);

    }

    public int readInt(ByteOrder bitOrder) throws IOException {

        return readInt(Integer.SIZE, bitOrder);

    }

    public int readInt(int len) throws IOException {

        return readInt(len, this.streamBitOrder);

    }

    public int readInt(int len, ByteOrder bitOrder) throws IOException {

        if (len == 0)

            return 0;

        if (len < 0 || len > Integer.SIZE)

            throw new IllegalArgumentException("Invalid len: " + len + " (must be 0-" + Integer.SIZE + ")");

        int value = 0;

        if (bitOrder == ByteOrder.BIG_ENDIAN) {

            for (int i = 0; i < len; i++)

                value = (value << 1) | readBit();

        } else {

            for (int i = 0; i < len; i++)

                value |= readBit() << i;

        }

        return value;

    }

}

测试

public static void main(String[] args) throws Exception {

    String bitData = "0101 00001 00001 00010 00011 00101 01000 01101 10101" // 5: 1, 1, 2, 3, 5, 8, 13, 21

                  + " 0011 000 001 010 011 100 101 110 111";                // 3: 0, 1, 2, 3, 4, 5, 6, 7

    BigInteger bi = new BigInteger(bitData.replaceAll(" ", ""), 2);

    System.out.println("0x" + bi.toString(16) + " = 0b" + bi.toString(2));

    byte[] byteData = bi.toByteArray();

    try (BitInputStream in = new BitInputStream(new ByteArrayInputStream(byteData))) {

        int[] nmbrs = readNmbrs(in);

        int[] nmbrs2 = readNmbrs(in);

        System.out.println(Arrays.toString(nmbrs));

        System.out.println(Arrays.toString(nmbrs2));

    }

}

private static int[] readNmbrs(BitInputStream in) throws IOException {

    int[] nmbrs = new int[8];

    int length = in.readInt(4);

    for (int i = 0; i < nmbrs.length; i++)

        nmbrs[i] = in.readInt(length);

    return nmbrs;

}

输出

0x5084432a1b53053977 = 0b10100001000010001000011001010100001101101010011000001010011100101110111

[1, 1, 2, 3, 5, 8, 13, 21]

[0, 1, 2, 3, 4, 5, 6, 7]