signed byte type and bitwise operators in Java?

Question

quoting from oracle website "byte: The byte data type is an 8-bit signed two's complement integer. It has a minimum value of -128 and a maximum value of 127 (inclusive)".

here, the first two lines are valid but the last is not

            byte b = -128;
        byte b1 = 127;
        byte b2 = b>>>b1;//illegal

Q1) what is meant exactly by 8-bit signed? 128 in binary format would be 1000 0000 and -128 would need an extra bit for the negative sign, where it would fit if all the 8 bits are occupied.

Q2) for int, there is a unsigned right shift operator, but that seems illegal with bytes, why is it so. can overflow not be prevented in case of bytes. In case of int, it works

Thanks for your help

rgettman rgettman · Accepted Answer · 2013-10-17T18:27:41

Just what it sounds like: There are 8 bits, holding 2^8 = 256 possible values. It's signed, so the range is frmo -128 through 127 (256 values). The most significant bit has a value of -128.
In Java, binary numeric promotion occurs with operations such as b >>> b1. Both types are promoted to int, and the result is an int. However, you can explicitly cast the result back to byte.

Here's the cast:

byte b2 = (byte) (b >>> b1);

The JLS, Section 5.6.2, talks about binary numeric promotion:

Widening primitive conversion (§5.1.2) is applied to convert either or both operands as specified by the following rules:

If either operand is of type double, the other is converted to double.

Otherwise, if either operand is of type float, the other is converted to float.

Otherwise, if either operand is of type long, the other is converted to long.

Otherwise, both operands are converted to type int.

(emphasis mine)

signed byte type and bitwise operators in Java?

1 Answers