基本类型与二进制转换

/**
 * Created by F.Fang on 2015/4/10.
 * Version :2015/4/10
 */
public final class ByteUtil {

    private ByteUtil(){}

    /**
     *
     * <pre>
     * 将4个byte数字组成的数组合并为一个float数.
     * </pre>
     *
     * @param arr
     * @return
     */
    public static float byte4ToFloat(byte[] arr) {
        if (arr == null || arr.length != 4) {
            throw new IllegalArgumentException("byte数组必须不为空,并且是4位!");
        }
        int i = byte4ToInt(arr);
        return Float.intBitsToFloat(i);
    }

    /**
     *
     * <pre>
     * 将一个float数字转换为4个byte数字组成的数组.
     * </pre>
     *
     * @param f
     * @return
     */
    public static byte[] floatToByte4(float f) {
        int i = Float.floatToIntBits(f);
        return intToByte4(i);
    }

    /**
     *
     * <pre>
     * 将八个byte数字组成的数组转换为一个double数字.
     * </pre>
     *
     * @param arr
     * @return
     */
    public static double byte8ToDouble(byte[] arr) {
        if (arr == null || arr.length != 8) {
            throw new IllegalArgumentException("byte数组必须不为空,并且是8位!");
        }
        long l = byte8ToLong(arr);
        return Double.longBitsToDouble(l);
    }

    /**
     *
     * <pre>
     * 将一个double数字转换为8个byte数字组成的数组.
     * </pre>
     *
     * @param i
     * @return
     */
    public static byte[] doubleToByte8(double i) {
        long j = Double.doubleToLongBits(i);
        return longToByte8(j);
    }

    /**
     *
     * <pre>
     * 将32位int转换为由四个8位byte数字.
     * </pre>
     *
     * @param sum
     * @return
     */
    public static byte[] intToByte4(int sum) {
        byte[] arr = new byte[4];
        arr[0] = (byte) (sum >> 24);
        arr[1] = (byte) (sum >> 16);
        arr[2] = (byte) (sum >> 8);
        arr[3] = (byte) (sum & 0xff);
        return arr;
    }

    /**
     *
     * <pre>
     * 将一个char字符转换为两个byte数字转换为的数组.
     * </pre>
     *
     * @param c
     * @return
     */
    public static byte[] charToByte2(char c) {
        byte[] arr = new byte[2];
        arr[0] = (byte) (c >> 8);
        arr[1] = (byte) (c & 0xff);
        return arr;
    }

    /**
     *
     * <pre>
     * 将2个byte数字组成的数组转换为一个char字符.
     * </pre>
     *
     * @param arr
     * @return
     */
    public static char byte2ToChar(byte[] arr) {
        if (arr == null || arr.length != 2) {
            throw new IllegalArgumentException("byte数组必须不为空,并且是2位!");
        }
        return (char) (((char) (arr[0] << 8)) | ((char) arr[1]));
    }


    /**
     * <pre>
     * 将长度为4的8位byte数组转换为32位int.
     * </pre>
     *
     * @param arr
     * @return
     */
    public static int byte4ToInt(byte[] arr) {
        if (arr == null || arr.length != 4) {
            throw new IllegalArgumentException("byte数组必须不为空,并且是4位!");
        }
        return (int) (((arr[0] & 0xff) << 24) | ((arr[1] & 0xff) << 16) | ((arr[2] & 0xff) << 8) | ((arr[3] & 0xff)));
    }

    /**
     *
     * <pre>
     * 将长度为8的8位byte数组转换为64位long.
     * </pre>
     *
     * 0xff对应16进制,f代表1111,0xff刚好是8位 byte[]
     * arr,byte[i]&0xff刚好满足一位byte计算,不会导致数据丢失. 如果是int计算. int[] arr,arr[i]&0xffff
     *
     * @param arr
     * @return
     */
    public static long byte8ToLong(byte[] arr) {
        if (arr == null || arr.length != 8) {
            throw new IllegalArgumentException("byte数组必须不为空,并且是8位!");
        }
        return (long) (((long) (arr[0] & 0xff) << 56) | ((long) (arr[1] & 0xff) << 48) | ((long) (arr[2] & 0xff) << 40)
                | ((long) (arr[3] & 0xff) << 32) | ((long) (arr[4] & 0xff) << 24)
                | ((long) (arr[5] & 0xff) << 16) | ((long) (arr[6] & 0xff) << 8) | ((long) (arr[7] & 0xff)));
    }

    /**
     * 将一个long数字转换为8个byte数组组成的数组.
     */
    public static byte[] longToByte8(long sum) {
        byte[] arr = new byte[8];
        arr[0] = (byte) (sum >> 56);
        arr[1] = (byte) (sum >> 48);
        arr[2] = (byte) (sum >> 40);
        arr[3] = (byte) (sum >> 32);
        arr[4] = (byte) (sum >> 24);
        arr[5] = (byte) (sum >> 16);
        arr[6] = (byte) (sum >> 8);
        arr[7] = (byte) (sum & 0xff);
        return arr;
    }

}