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 · Accepted Answer · 2013-10-17 18:27:41Z

5

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)

answered Oct 17, 2013 at 18:27

rgettman

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8 Comments

eagertoLearn Over a year ago

thanks, in terms of 8-bits, where does the sign go? + and - should be part of 8-bit right?

rgettman Over a year ago

Think of the value of the byte as the sum of the values of all 8 bits: -128, 64, 32, 16, 8, 4, 2, and/or 1.

eagertoLearn Over a year ago

That means a 8-bit representation of 1000 0000 should be -128? and not 128.?

rgettman Over a year ago

Right, -128 is represented as 1000 0000, and 127 is represented as 0111 1111. That also means that -1 is represented as 1111 1111.

Stijn de Witt Over a year ago

'-128' is '1000 0000', from there you count back to zero like this: '-127' is '1000 0001', '-126' = '1000 0010', '-125' = '1000 0011', .. '-1' is '1111 1111', '0' is '0000 0000'

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