Unsigned right shift

[Kezia Matthews]

Hi Everybody,
Can anybody explain the output of the following program for me?

class UnsignedRightShift
{
public static void main (String args[])
{
int i = -1;
System.out.println (i>>>32);
}
}

The output is -1.
For unsigned right shift, since sign is not honoured, zeros are filled in the vacancies created in the MSB position for each shift to the right. If I am not wrong, since the bit size of int is 32, after 32 shifts, all the 32 bits would be replaced by zeros. The output should've been zero. I think that my understanding of this is wrong. Can anybody help me with some explanation?
Thanks.
Kezia.

[Dave Vick]

Kezia
Your understanding is right however, before it performs a shift the java performs a bitwise & on the right hand operand with a mask of 0x1F for ints and 0x3F for longs. For a positive number this is equivalent to doing a %32 or %64.
So, if you're right hand operand is 32, then 32%32 is 0 and no shift is performed.
If you want a detailed explaination check out the JLS section 15.19
for an easier to see example try to do a shift of 33 or 34 and you see that the shift is actually 1 or 2.
hope that helps

------------------
Dave
Sun Certified Programmer for the Java� 2 Platform

[Kezia Matthews]

Thanks Dave for the quick reply.
Kezia.