Unsigned Right Shift operator works as the normal right-shift operator, why?-CodePudding

When I use the Unsigned Right Shift (Logical Shift) operator in java, I get the same result as the normal right-shift operator would give. my code is:

        byte b1 = -99; // 01100011
        byte result = (byte) (b1 >>> 6);
        String  str = String.format("%8s", Integer.toBinaryString(result & 0xFF)).replace(' ', '0');
        System.out.println("result: "   result);

The preceding code produces the same output which is -13 with both of >> and >>> operators, what is the reason behind that and how to resolve it?

CodePudding user response：

First of all, byte is a signed type in Java, and bit shifting (whether normal or with sign extension) is done in int, so before performing the shift, the value of b1 is implicitly cast back to int (which preserves sign). (As an aside, the value -99 is not 01100011, it is 1001 1101 in byte and 1111 1111 1111 1111 1111 1111 1001 1101 in int).

With b >> 6 (or -99 >> 6), the value is -2 (1111 1111 1111 1111 1111 1111 1111 1110), with b >>> 6 (or -99 >>> 6) the value in int is 67108862 (0000 0011 1111 1111 1111 1111 1111 1110). However, when you then cast to byte, this last value becomes -2 as well (1111 1110).

What you need to do is make sure you get the unsigned value of the byte as the integer value before shifting:

byte result = (byte) ((b1 & 0xFF) >>> 6);

Which results in 2.