How to get a 42 bit long number in Java.

[Ashish Schottky]

Java's long comprises of 63 bits where in the MSB is stored for the sign.

I want to set 42nd bit of a particular number as 1.
So I tried 1<<42, however this doesnt give the output as what I thought.
neither did this work 1<<(42L).
I am not supposed to use Strings and other wrapper class functions.
Is there anyway by which I can fix this up? This is a small part of my program but a vital one.
I am getting the answer to be 1024, this very well means that it is treating it to be a 32 bit number and then 1<<10.
confused with this.
-Thank you.

[fred rosenberger]

What did you get, and what did you expect? and please post your actual code (in code tags) so we can see what you are REALLY doing.

I have a few guesses as to what the problem may be, but without seeing what you're doing, and knowing what you are expecting, they would only be guesses.

edit: I believe the problem is with your '1' literal. try declaring it as a long :

1L << 42

also.... the set bit is already in the first position, so you may need to only shift it 41.

[Ashish Schottky]

@Fred
Thanks for reply.
Well I am writing this independently and once I get this part working I will integrate it with my code.

long bitboard=(1<<(42));
        System.out.println(bitboard);
   bitboard=(1<<(42L));
        System.out.println(bitboard);
   

I was expecting 4398046511104
where as I got 1024.
Edit after you edited:
Thanks a lot it solved my problem.

[Paul Clapham]

I was going to make a suggestion, but I see Fred has already made that suggestion. So now my suggestion is to read and implement Fred's suggestion.

[fred rosenberger]

Ashish Schottky wrote:
Thanks a lot it solved my problem.

More importantly, do you understand WHY it solved your problem?