shift operator

[Basanti Mathad]

Hey,
can any one explain the step wise calculation for this.I don't understand why the result is -1.
--------------------------------------------
class J {
public static void main (String[] s) {
byte b = 0;
b += ~b >>> 1;
System.out.println(b);
}
}
---------------------------------------------
Thank u.

[Michael Ernest]

I imagine this code is intended to test your understanding of signed 2's-complement behavior as well as operator precedence.
Which comes first the unsigned shift operator or the bitwise operator? It's not too baffling, really -- just find the table. Or, just demonstrate it to yourself:

public class Bit
{
  public static void main(String args[])
  {
    int b = 0;
    b = ~b;
    System.out.println("The bit-complement of zero is: " + b);
    b = b >>> 1;
    System.out.println("The right-shift of the complement is: " + b);
    b = 0;
    b = b >>> 1;
    System.out.println("The right-shift of zero is: " + b);
    b = ~b;
    System.out.println("The bit-complement of the right-shift is: " + b);
  }
}

The real test of this code, in my opinion, is to see how quickly you'll think to write your own demonstration.
Elsewhere in another forum a poster was commenting that it seemed silly to test an SCJP with a technical interview. I couldn't disagree more. If I have learned anything from this site, it is that some people are astoundingly good test takers. What I want to know when I hire someone, however, is not whether they're proficient in the elements of the language but how they work. I'd sooner hire someone who can show their work than someone who can rattle everything off from memory.
Do the work! It seems soooo slow to do things this way but you'll get faster. Write code any time you have a doubt; you'll learn far more.

[John Lee]

Originally posted by Basanti Mathad:
Hey,
can any one explain the step wise calculation for this.I don't understand why the result is -1.
--------------------------------------------
class J {
public static void main (String[] s) {
byte b = 0;
b += ~b >>> 1;
System.out.println(b);
}
}
---------------------------------------------
Thank u.

byte b = 0;
b = 00000000;
~b = 11111111;
~b >>> 1 = 11111111; (here, ~b is promoted to int first, then truncated to byte)
b += ~b >>> 1;
b = 11111111 ;
b = -1; (here, byte range is from -2^7 to 2^7 - 1, 11111111 add up to 2^8 - 1, so must be '-')

[sam huang]

Try this:

 class Bit {
public static void main (String[] s) {
byte b = 0;
b =(byte) ( b + ~b>>>1 ) ;
//b += ~b>>>1;
System.out.println(b);
}
} 

[sam huang]

I think:
~b>>>1 = 2147483647 (01111111111111111111111111111111)

[Basanti Mathad]

Thank u all guys........
i was confused bcoz it had assignment operator.

[Prosenjit Banerjee]

I can explain it in this way:
2's complent of x means 1's complement + 1 i.e.
2's comp of x = ~x + 1
Again, a -x is nothing but 2's complement of x.
So, -x = ~x + 1
Thus, remember ~x = -x - 1
Now bit pattern of 0 is nothing but a string of 0s -> 000...000
So, bit pattern of ~0 will be surely a string of 1s -> 111...111
Signed right shift operator >>> makes no change to this pattern 111...111
because, this operator fills the left bits(s) with the left most bit of the
original pattern (1 in this case). So, this is also as same as ~0 which is
nothing but -0 - 1 = -1 (as stated before. -1 added to 0 becomes -1.
Please specify if I am wrong.