Java boolean[] 到 byte[] 并返回

Question

我通过 Java 中的套接字连接发送 byte[] 数组。

我有一个很长的boolean[] 数组，其中array.length % 8 == 0。

我想将这个boolean[] 数组转换为一个元素少8 倍的byte[] 数组，这样我就可以通过套接字连接发送byte[]。

boolean[] 数组如下所示：01011010 10101010 01100011 11001010 等

在这种情况下，byte[] 应如下所示：0x5A 0xAA 0x63 0xCA。

我在另一个问题上找到了一些代码，关于如何将单个 byte 转换为 boolean[] 数组，并在此处添加了一个新方法来转换整个数组：

public static boolean[] booleanArrayFromByteArray(byte[] x) {
    boolean[] y = new boolean[x.length * 8];
    int position = 0;
    for(byte z : x) {
        boolean[] temp = booleanArrayFromByte(z);
        System.arraycopy(temp, 0, y, position, 8);
        position += 8;
    }
    return y;
}

public static boolean[] booleanArrayFromByte(byte x) {
    boolean bs[] = new boolean[4];
    bs[0] = ((x & 0x01) != 0);
    bs[1] = ((x & 0x02) != 0);
    bs[2] = ((x & 0x04) != 0);
    bs[3] = ((x & 0x08) != 0);
    return bs;
}

我想知道是否有更有效的方法。

编辑：谢谢

Answer 1

从字节转换为布尔值：

private static boolean[] convertToBooleanArray(byte[] bytes) {
    boolean [] result = new boolean[bytes.length * 8];

    for (int i=0; i<bytes.length; i++) {
        int index = i*8;
        result[index+0] = (bytes[i] & 0x80) != 0;
        result[index+1] = (bytes[i] & 0x40) != 0;
        result[index+2] = (bytes[i] & 0x20) != 0;
        result[index+3] = (bytes[i] & 0x10) != 0;
        result[index+4] = (bytes[i] & 0x8) != 0;
        result[index+5] = (bytes[i] & 0x4) != 0;
        result[index+6] = (bytes[i] & 0x2) != 0;
        result[index+7] = (bytes[i] & 0x1) != 0;
    }

    return result;
}

通过切断最后几个布尔值将布尔值转换为字节：

/**
 * This will round down to the nearest number of bytes.  So it will chop off the last few booleans.
 * Eg: If there are 9 booleans, then that will be 1 byte, and it will lose the last boolean.
 */
private static byte[] convertToByteArray(boolean[] booleans) {
    byte[] result = new byte[booleans.length/8];

    for (int i=0; i<result.length; i++) {
        int index = i*8;
        byte b = (byte)(
                (booleans[index+0] ? 1<<7 : 0) +
                (booleans[index+1] ? 1<<6 : 0) +
                (booleans[index+2] ? 1<<5 : 0) + 
                (booleans[index+3] ? 1<<4 : 0) +
                (booleans[index+4] ? 1<<3 : 0) +
                (booleans[index+5] ? 1<<2 : 0) + 
                (booleans[index+6] ? 1<<1 : 0) +
                (booleans[index+7] ? 1 : 0));
        result[i] = b;
    }

    return result;
}

通过填充最后几个布尔值将布尔值转换为字节：

/**
 * This will pad to the nearest number of bytes.  So the last few booleans will be set to the padValue.
 * Eg: If there are 9 booleans, then the last 7 will be added as the padValue (making 16 booleans).
 * 
 * @param booleans
 * @param padValue
 * @return
 */
public static byte[] convertToByteArray(boolean[] booleans, boolean padValue) {
    boolean[] paddedBooleans;
    int remainder = booleans.length % 8;

    // Booleans are already divisible by 8, nothing to pad
    if (remainder == 0) {
        paddedBooleans = booleans;
    }
    // Boolean are not divisible by 8, need to pad
    else {
        int padAmount = 8 - remainder;
        paddedBooleans = Arrays.copyOf(booleans, booleans.length + padAmount);

        for (int i=booleans.length; i<paddedBooleans.length; i++) {
            paddedBooleans[i] = padValue;
        }
    }

    // Convert the boolean array into a byte array
    byte[] result = new byte[paddedBooleans.length/8];

    for (int i=0; i<result.length; i++) {
        int index = i*8;
        byte b = (byte)(
                (paddedBooleans[index+0] ? 1<<7 : 0) +
                (paddedBooleans[index+1] ? 1<<6 : 0) +
                (paddedBooleans[index+2] ? 1<<5 : 0) + 
                (paddedBooleans[index+3] ? 1<<4 : 0) +
                (paddedBooleans[index+4] ? 1<<3 : 0) +
                (paddedBooleans[index+5] ? 1<<2 : 0) + 
                (paddedBooleans[index+6] ? 1<<1 : 0) +
                (paddedBooleans[index+7] ? 1 : 0));
        result[i] = b;
    }

    return result;
}

Answer 2

标准库中已经存在一个类 (BitSet) 来帮助您执行此操作，您应该使用它而不是布尔数组。该类允许您获取和设置位以及对布尔值作为一个组进行各种逻辑操作。

例如。

BitSet bits = BitSet.valueOf(bytes); 
boolean third_bit = bits.get(3);
bits.set(5, false);
byte[] new_bytes = bits.toByteArray();

如果您确实需要使用布尔数组，那么以下方法将起作用。

static boolean[] toBooleanArray(byte[] bytes) {
    BitSet bits = BitSet.valueOf(bytes);
    boolean[] bools = new boolean[bytes.length * 8];
    for (int i = bits.nextSetBit(0); i != -1; i = bits.nextSetBit(i+1)) {
        bools[i] = true;
    }
    return bools;
}

static byte[] toByteArray(boolean[] bools) {
    BitSet bits = new BitSet(bools.length);
    for (int i = 0; i < bools.length; i++) {
        if (bools[i]) {
            bits.set(i);
        }
    }

    byte[] bytes = bits.toByteArray();
    if (bytes.length * 8 >= bools.length) {
        return bytes;
    } else {
        return Arrays.copyOf(bytes, bools.length / 8 + (bools.length % 8 == 0 ? 0 : 1));
    }
}

Answer 3

用于将字节转换为布尔值：

public static boolean [] bytesToBooleans(byte [] bytes){
    boolean [] bools = new boolean[bytes.length * 8];
    byte [] pos = new byte[]{(byte)0x80, 0x40, 0x20, 0x10, 0x8, 0x4, 0x2, 0x1};

    for(int i = 0; i < bytes.length; i++){
        for(int j = i * 8, k = 0; k < 8; j++, k++){
            bools[j] = (bytes[i] & pos[k]) != 0;
        }
    }

    return bools;
}

或者其他方法：

public static boolean [] bytesToBooleansAnotherWay(byte [] bytes){
    boolean [] bools = new boolean[bytes.length * 8];

    for(int i = 0; i < bytes.length; i++){
        int j = i * 8;
        bools[j] = (bytes[i] & 0x80) != 0;
        bools[j + 1] = (bytes[i] & 0x40) != 0;
        bools[j + 2] = (bytes[i] & 0x20) != 0;
        bools[j + 3] = (bytes[i] & 0x10) != 0;
        bools[j + 4] = (bytes[i] & 0x8) != 0;
        bools[j + 5] = (bytes[i] & 0x4) != 0;
        bools[j + 6] = (bytes[i] & 0x2) != 0;
        bools[j + 7] = (bytes[i] & 0x1) != 0;
    }

    return bools;
}

Answer 4

你可以这样做。

public byte[] toBytes(boolean[] input) {
    byte[] toReturn = new byte[input.length / 8];
    for (int entry = 0; entry < toReturn.length; entry++) {
        for (int bit = 0; bit < 8; bit++) {
            if (input[entry * 8 + bit]) {
                toReturn[entry] |= (128 >> bit);
            }
        }
    }

    return toReturn;
} 

这依赖于toReturn 将被初始化为全零的事实。然后对于我们在input 中遇到的每个true，我们在toReturn 的相应条目中设置相应的位。