adding byte values question...

[Jay Ashar]

public static void main( String[] args )
{
byte b = 127;
byte c = (byte)(b + 1);
System.out.println(b);
System.out.println(c);
}
When I compile and run this, it displays c as -128. Why is it doing this?
byte b is 127 which is 01111111
how does adding 1 to it gives 10000000.
I am bit confused on this, appreciate your help.
Thanks in advance.

[Corey McGlone]

If you add 00000001 to 01111111, you get this:

  00000001
+ 01111111
----------
  10000000

Of course, in Java, the leftmost bit is a sign bit. It only tells you if a number is positive or negative. If it's a 0, the number is positive and, if it's a 1, the number is negative. Therefore, our addition is rolling us over from the maxiumum value you can contain in a byte (127) to the minimum value you can contain in a byte (-128).
If you do a search in this forum on 2's complement, I'm sure you can find a lot of good descriptions of this.

[Bojan Knezovic]

Originally posted by Jay Patel:
public static void main( String[] args )
{
byte b = 127;
byte c = (byte)(b + 1);
System.out.println(b);
System.out.println(c);
}
When I compile and run this, it displays c as -128. Why is it doing this?
byte b is 127 which is 01111111
how does adding 1 to it gives 10000000.
I am bit confused on this, appreciate your help.
Thanks in advance.

You're getting overflow, bytes can store numbers from -128 to 127.
127=0111 1111
0111 1111
+0000 0001
----------
1000 0000

To read the number apply the 2's complement rules:
1) complement: ~1000 0000 = 0111 1111
2) add 1: 0111 1111 + 1 = 1000 0000
3) read the number: 1000 0000 = 128
4) add the sign and you get: -128.

HTH,
Bojan
[ February 25, 2004: Message edited by: Bojan Knezovic ]

[Jay Ashar]

Thanks a lot you'll