Question

这是

Long number = 0x001122334455667788L;

我需要创建Long的最后6个字节的byte[]。

所以它看起来像

byte[] bytes = {0x22,0x33,0x44,0x55,0x66,0x77,0x88};

制作这样的东西的正确方法是什么？

感谢任何回应

Answer 1

java.lang.BigInteger toByteArray()

Answer 2

    byte[] buffer = new byte[6];
    buffer[0] = (byte)(v >>> 40);
    buffer[1] = (byte)(v >>> 32);
    buffer[2] = (byte)(v >>> 24);
    buffer[3] = (byte)(v >>> 16);
    buffer[4] = (byte)(v >>>  8);
    buffer[5] = (byte)(v >>>  0);

这就是DataOutputStream.writeLong()的作用（除了它写了所有的字节，或者当然）

Answer 3

如何使用DataOutputStream？

ByteArrayOutputStream baos = new ByteArrayOutputStream(); // This will be handy.
DataOutputStream os = new DataOutputStream(baos);
try {
  os.writeLong(number); // Write our number.
} catch (IOException e) {
  e.printStackTrace();
} finally {
  try {
    os.close(); // close it.
  } catch (IOException e) {
    e.printStackTrace();
  }
}
return Arrays.copyOfRange(baos.toByteArray(), 2, 8); // Copy out the last 6 elements.

Answer 4

您可以使用ByteBuffer

Long number = 0x001122334455667788L;
ByteBuffer buffer = ByteBuffer.allocate(8);
buffer.putLong(number);
byte[] full = buffer.array();       
byte[] shorter = Arrays.copyOfRange(full, 2, 8); // get only the lower 6

Answer 5

BigInteger也会这样做。

BigInteger number = new BigInteger("001122334455667788", 16);
byte[] b = number.toByteArray();
// May need to tweak the `b.length - 6` if the number is less than 6 bytes long.
byte[] shortened = Arrays.copyOfRange(b, b.length - 6, b.length);
System.out.println(Arrays.toString(shortened));

从类型long获取字节的最快方法

5 个答案: