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一个int占用4个byte，所以需要声明一个4位长度的byte数组

public byte[] int2Byte2(int i) {
        byte[] bytes = new byte[4];
        bytes[0] = (byte) ((i >> 24) & 0xff);
        bytes[1] = (byte) ((i >> 16) & 0xff);
        bytes[2] = (byte) ((i >> 8) & 0xff);
        bytes[3] = (byte) (i & 0xff);
        return bytes;
    }

0xff：0x表示16进制，后边的ff表示15*15=255，也就是1111 1111，当进行&操作时，会先将自己补位（变为int类型），也就是 0000 0000 0000 0000 0000 0000 1111 1111

&操作规则：https://blog.csdn.net/m0_47759414/article/details/107148516

这里是先进行移位再进行&操作，最后进行强转byte，此处我认为进行&操作是没必要的，因为&操作的主要目的就是将低八位以外的位置进行置0操作，但是进行强制类型转换byte后，只需要保留低八位，那么其余位置的数据是有没有置0，又有什么关系呢。

但是看到了一块apache的源码进行转换时，也进行了&操作，所以到底什么场景会用到&操作呢？

    /**
     * Returns the integer represented by up to 4 bytes in network byte order.
     * 
     * @param buf the buffer to read the bytes from
     * @param start
     * @param count
     * @return
     */
    public static int networkByteOrderToInt(byte[] buf, int start, int count) {
        if (count > 4) {
            throw new IllegalArgumentException("Cannot handle more than 4 bytes");
        }
 
        int result = 0;
 
        for (int i = 0; i < count; i++) {
            result <<= 8;
            result |= (buf[start + i] & 0xff);
        }
 
        return result;
    }
 
    /**
     * Encodes an integer into up to 4 bytes in network byte order.
     * 
     * @param num the int to convert to a byte array
     * @param count the number of reserved bytes for the write operation
     * @return the resulting byte array
     */
    public static byte[] intToNetworkByteOrder(int num, int count) {
        byte[] buf = new byte[count];
        intToNetworkByteOrder(num, buf, 0, count);
 
        return buf;
    }
 
    /**
     * Encodes an integer into up to 4 bytes in network byte order in the 
     * supplied buffer starting at <code>start</code> offset and writing
     * <code>count</code> bytes.
     * 
     * @param num the int to convert to a byte array
     * @param buf the buffer to write the bytes to
     * @param start the offset from beginning for the write operation
     * @param count the number of reserved bytes for the write operation
     */
    public static void intToNetworkByteOrder(int num, byte[] buf, int start, int count) {
        if (count > 4) {
            throw new IllegalArgumentException("Cannot handle more than 4 bytes");
        }
 
        for (int i = count - 1; i >= 0; i--) {
            buf[start + i] = (byte) (num & 0xff);
            num >>>= 8;
        }
    }