A question ...

[Ira Jain]

Hi,
Can somebody please explain me the output of the following question-
class MyClass{
public static void test(int i, String exp) {
if((i>>1)!=(i>>>1)) System.out.println(exp);
}
public static void main(String[] args){
test(1<<32,"1<<32");
test(1<<31,"1<<31");
test(1<<30,"1<<30");
test(1,"1");
test(0,"0");
test(-1,"-1");
}
}
The given ouput is : 1<<31 and -1.
thanx in advance
Ira

[Paul Anilprem]

This is actually quite simple. The point is, >> and >>> produce the same resule EXCEPT for the -ive integers. (And of course, if the RHS is not a multiple of 32/64)
Here, only two are -ive 1 << 31 and -1! Hence, the answer.<br /> 1<< 31, puts 1 on the 32nd bit which is a sign bit making it -ive.<br /> I hope you know why >> and >>> produce different result for -ive integers.
HTH,
Paul.

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[This message has been edited by Paul Anil (edited October 17, 2000).]

[bill bozeman]

Paul is right, and just in case someone doesn't know why >> and >>> produce different answers for negative numbers, it is because >> looks at the highest bit value and that is what it adds to the beginning of the bit pattern, and >>> always adds 0's. Since negative numbers have a 1 at the highest bit value, they stay negative, while >>> doesn't.
[This message has been edited by bill bozeman (edited October 17, 2000).]

[Ira Jain]

Thanks a lot Bill and Paul for clearing my doubts.