Question

This seems fairly straightforward, but I cant find an answer. If I have an int X, what is the best way to get N least significant bits from this int, in Java?

Answer 1

This should work for all non-negative N < 33 32:

x & ((1 << N) - 1)

It’s worth elaborating on how this works for N == 31 and N == 32. For N == 31, we get 1 << N == Integer.MIN_VALUE. When you subtract 1 from that, Java silently wraps around to Integer.MAX_VALUE, which is exactly what you need. For N == 32, the 1 bit is shifted completely out, so 1 << N == 0; then (1 << N) - 1 == -1, which is all 32 bits set.

For N == 32, this unfortunately doesn’t work because (thanks, @zstring!) the << operator only shifts by the right side mod 32. Instead, if you want to avoid testing for that case specially, you could use:

x & ((int)(1L << N) - 1)

By shifting a long, you get the full 32-bit shift, which, after casting back to an int, gets you 0. Subtracting 1 gives you -1 and x & -1 is just x for any int value x (and x is the value of the lower 32 bits of x).