If you mean Shift-Right-With-Zero-Fill, the >>> operator, then that's probably a trick question. If you shift an integer 0 or any multiple of 32 bits to the right, you will get the same number again. Take this example:
int x = -1;
x = x >>> 0; // x is still -1
x = x >>> 32 // x is still -1
And so on. Now honestly I'm not sure why you get back to the same number after 32, unless it somehow subtracts the highest multiple of 32 from the bit shift number, then shifts??? Dangit, now I have a question...
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Michael J Bruesch
Codito, ergo sum...
I code, therefore I am. http://www.geocities.com/mjbruesch