Java boolean []到byte []并返回 [英] Java boolean[] to byte[] and back

问题描述

我正在通过Java中的套接字连接发送byte[]数组.

I am sending byte[] arrays over a socket connection in Java.

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

I have a pretty long boolean[] array, where array.length % 8 == 0.

我想将此boolean[]数组转换为元素数量减少8倍的byte[]数组，以便随后可以通过套接字连接发送byte[].

I'd like to convert this boolean[] array into a byte[] array with 8 times as few elements, so that I can then send the byte[] over the socket connection.

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

在这种情况下，byte[]应该看起来像这样:0x5A 0xAA 0x63 0xCA.

The byte[] in this case should look like this: 0x5A 0xAA 0x63 0xCA.

我在另一个问题上找到了一些有关如何将单个byte转换为boolean[]数组的代码，并在此处添加了新方法以转换整个数组:

I have found some code on another question on how to convert a single byte into a boolean[] array and added a new method to it to convert an entire array here:

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.

I'd like to know if there is a more efficient way of doing this.

谢谢

推荐答案

您可以这样做.

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将被全零初始化的事实.然后针对在input中遇到的每个true，在toReturn的相应条目内设置相应的位.

This relies on the fact that toReturn will be initialised with all zeroes. Then for each true that we encounter in input, we set the appropriate bit within the appropriate entry in toReturn.

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