negative numbers

[Paul Ze]

Here is yet another question from the Sun Learning center practice exam site.
Given:
1. public class juju {
2. public static void main(String [] args) {
3. double z = 0xFFFFFFFC;
4. int i = 0xFFFFFFDF;
5. double x = z * i;
6. System.out.println(x);
7. }
8. }
What is the result?
A
Compilation fails because of an error on line 3.
B
Compilation fails because of an error on line 5.
C
Compilation succeeds and the program prints "-37.0".
D
Compilation succeeds and the program prints "132.0".
The answer is D because both of those numbers are negative numbers. But how do i know that those two numbers are negative from the reading the code?

[Valentin Crettaz]

negative numbers always have the sign bit (the left most bit) set to 1.
z is FFFFFFFC which translates to binary to 11111111 11111111 11111111 11111100
i is FFFFFFDF which translates to binary to 11111111 11111111 11111111 11011111
Now, z and i obviously have their left-most (sign) bit set to 1, so they are both negative numbers.
The rest, namely how you get to 132, is just a matter of translating z and i to decimal and multiplying them:
z = 0xFFFFFFFC = 11111111 11111111 11111111 11111100 = -4
i = 0xFFFFFFDF = 11111111 11111111 11111111 11011111 = -33
z*i = -4 * -33 = 132
[ March 16, 2002: Message edited by: Valentin Crettaz ]

[Rajinder Yadav]

learn about hexadecimal number, basically you're looking to see if the most significant bit has been set, if this bit is a (binary) '1', then the number is negative, otherwise it's a positive number
[ March 16, 2002: Message edited by: Rajinder Yadav ]

[Dirk Schreckmann]

Anybody have a link to learn about the hexidecimal number format?

[Valentin Crettaz]

Google > Hexadecimal ?
http://www.webopedia.com/TERM/H/hexadecimal.html
http://www.cs.ualberta.ca/~casey/c101/101notes/hex.html
http://www.jaworski.com/htmlbook/dec-hex.htm
http://www.mindprod.com/jglossh.html#HEX
...
Also, a search in this forum may help

[Mark Pinks]

Originally posted by Valentin Crettaz:
negative numbers always have the sign bit (the left most bit) set to 1.
z is FFFFFFFC which translates to binary to 11111111 11111111 11111111 11111100
i is FFFFFFDF which translates to binary to 11111111 11111111 11111111 11011111
Now, z and i obviously have their left-most (sign) bit set to 1, so they are both negative numbers.
The rest, namely how you get to 132, is just a matter of translating z and i to decimal and multiplying them:
z = 0xFFFFFFFC = 11111111 11111111 11111111 11111100 = -4
i = 0xFFFFFFDF = 11111111 11111111 11111111 11011111 = -33
z*i = -4 * -33 = 132
[ March 16, 2002: Message edited by: Valentin Crettaz ]

Hi, this is my first post to JavaRanch!
I'm a bit confused how 0xFFFFFFFC = -4
I thought you converted FFFFFFFC to dec as follows:
15*(16^7) + 15*(16^6) + 15*(16^5) + 15*(16^4) + 15*(16*3) + 15*(16^2) + 15*(16^1) + 12*(16^0) = 4294967292
Is it something to do with converting it to binary (1111111111111111111111111111100), removing the first bit (the negative sign), and then doing something else?

[raman padma]

Hi,
Here is the short cut :
~x = -x - 1
In other words

-x = ~x + 1

Even if they give a big number, do not worry about the entire bit operation. For ex double d = 0xFFFFFFFFC

By seeing the letter F, u know the number is negative. Now worry about the last byte (i.e) C
C = 12 (ie) x = 1101
~x = 0010
1 +
------
0011
--------
= - 3
I hope i am clear.
One more trick, By looking at ur example number u know that both the numbers are negative. If u multiply two negative numbers , u get a positive number. Simple the obvious choice is 132.
Regards,
Radha

[Mark Pinks]

Thanks a lot. I get it now.
But in your example,

C = 12 (ie) x = 1101
~x = 0010
1 +
------
0011
--------
= - 3

...did you really mean:
C = 12 (ie) x = 1100
~x = 0011
1 +
------
0100
--------
= - 4
?

[raman padma]

Hi,
OOOPS. Sorry for the mistake. u r right.
Regards,
Radha Shankar