# help?

feng lee
Ranch Hand
Posts: 33
public static void main(String args[])
{
int t=0xffff1;
int j=~t;
System.out.println(j);
}
What will be the output?
A. -1048562
B. -10
C. Exception
D. Compilation Error
Ans:A
why?

david eberhardt
Ranch Hand
Posts: 158

steps:
1) convert Oxffff1 to a 32 bit binary number
working backwards from above
1 = 0001
f = 1111 (which is 15 in decimal)
f = 1111
f = 1111
f = 1111
is 1111 1111 1111 1111 0001
that's a total of 20 bits so far.
since this is a positive number, we will add enough zeroes to total 32 bits:
0000 0000 0000 1111 1111 1111 1111 0001
------------------------------------------------
2) now we apply the ~ operator which flips each bit to it's opposite:
1111 1111 1111 0000 0000 0000 0000 1110
so now this is stored in the computer.
-------------------------------------------------
3) when we print this number it will be a negative number displayed because the highest bit (to the left above) is a 1
4) now we will attempt to print out this number, so two's compliment conversion is applied:
a) we apply the ~ operator again:
0000 0000 0000 1111 1111 1111 1111 0001
b) add 1 to above and get
0000 0000 0000 1111 1111 1111 1111 0010
now we convert this binary to decimal:
the leftmost 1 above is going to yield 2^19 = 524,288
the next 1 gives us 2^18 = 262,144
and so on ....
remember that when we get to the last 4 binary digits (0010) we have added up all these numbers as we go, for a total of 1048560 to this point:
the last four positions yield
2^3 *0 = 0 // 2^3 = 8
2^2 *0 = 0 // 2^2 = 4
2^1 *1 = 2 // 2^1 = 2
2^0 *0 = 0 // 2^0 = 1
So 1048560 + 2 = 10485562
And finally we put the negative sign in front to get -10485562 when we println(j)
note: int t was 1048561
and int j is -1048562

(the number 0 counts as a position on the number line)

david eberhardt
Ranch Hand
Posts: 158
oh, I know that was alot of work ... but I wanted to show evey detail.
Once we converted the hex number 0xffff1 to binary, we knew we had a big number because the leftmost 1 was 2^19 = 524,288 plus all the others...
then we were going to ~ that number turning it to some big negative number so that eliminated the answer -10.
next, we could eliminate the exception answer because we would not exceed the max value for an int at runtime!
compilation error is elimated because we didn't have to worry about casting and other things ... the code was written properly in all other aspects.
we just have to remember the +1 part when doing the two's compliment routine when we bring a negative number out to print it.
Hope this helps!

feng lee
Ranch Hand
Posts: 33
Thanks very much,david eberhardt.But i have another question.if t is -1048561,must we write
like this Ox800ffff1,thanks again!

feng lee
Ranch Hand
Posts: 33
I know it should write like this:
0xfff0000f

John Paverd
Ranch Hand
Posts: 115
System.out.println(Integer.toHexString(-1048562));
fff0000e