CODEWITHSUNDEEP
javabit-manipulation
user_1357
user_1357

Reputation: 7940

Setting 32nd bit for Java long type yields -1

I am using Java's primitive long data type to store a flag. long has 8 bytes so 64 bit with high order bit representing the sign bit so we have 63 usable bit as a flag. Everything works until I turn on 32nd bit yields -1. Why is this happening?

Edit: Basically I am trying to flag the availability in an hour for first 60 bit (1 bit representing a minutes in an hour) on long data type. Let's assume start time is 0:0:00. Everything works if the end time is up to 0:30:00 but fails with 0:31:00 up to 60 minutes, meaning when I attempt to (value & (1 << 31)) for some value initialized to 0 this will not return 2^31 rather it returns -1.

Edit: Declaring a mask as 1L rather than 1 solved the problem.

Upvotes: 0

Views: 226

Answers (3)

Stephen C
Stephen C

Reputation: 719596

I think this is your bug:

(1 << mIndex)

That is performing a shift on a 1 which is an int. If you perform a shift on an int, the result will be an int, even if the 2nd operand is greater than 32.

Try this instead:

(1L << mIndex)

Upvotes: 3

Fritz H
Fritz H

Reputation: 3569

Agreed with Lawrence Kesteloot, but if you still have no luck this may be a bug in your JVM implementation. You may also have better luck using the java.util.BitSet class.

Upvotes: -1

Lawrence Kesteloot
Lawrence Kesteloot

Reputation: 4398

You should paste your code, but my guess is that you've got an "int" somewhere in your math, and that's getting turned into -1 before it gets cast to a long.

Upvotes: 3

Related Questions