Unsigned Shift Operator（>>>）并将int转换为Java中的字节数组

Question

I want to convert integer to byte[] in java. And I did some research and found the code

    public byte[] integerToBytes (int i) {

          byte[] result = new byte[4];
          result[0] = (byte) (i >>> 24);

          result[1] = (byte) (i >>> 16);

          result[2] = (byte) (i >>> 8);

          result[3] = (byte) (i /*>> 0*/);

          return result;
    }

I'm pretty sure the code is right cause it passes the tests. However, I'm a little bit confused.

Suppose the integer is 36666666. And binary representation is 10001011110111110100101010.I can understand why it is true for result[0] since after the shift, it becomes 10001011(0x22).However,for result[1],after the shift,it becomes 1000101111011111(0x22F7).

But what I really want is just 0xF7.

Can someone explain this to me or I understand this code in a wrong way?

Cheers

Answer 1

The reason result[1] and others are just the lowest eight bits is due to the the cast to a byte performed on each assignment. Casting an int like 0x22F7 to a byte cuts off the top 3 bytes and leaves the least significant byte to be assigned to the new variable. Thus 0x22f7 becomes just 0xf7. Hope this helps.