Java gets a range of bits from long

Question

I need to extract certain bit ranges with a long value, for example:

long input = 15367 (11110000000111) 

What I need to do is extract two long values ​​from the original long,

 First long is 5 bits starting from bit 0, so bits 0:4 = 7 (0111) Second long is 56 bits starting from bit 8, so bits 7:55 = 60 (1111000) 

I know that this can be done using bit shifting and masking, however, I'm not quite sure how to implement this so that it is dynamic every time, since every time I need to do this, the long ones will be different, and therefore will also be specific bit ranges.

I read about BitSets and BitArrays, however I'm not quite sure if these are the right things to work.

Any advice on the best way to implement it is welcome.

Thanks!

Answer 1

To extract the nrBits bits starting at offset offset , you can do:

 public static long extractSub(final long l, final int nrBits, final int offset) { final long rightShifted = l >>> offset; final long mask = (1L << nrBits) - 1L; return rightShifted & mask; } 

Pay attention to the user of the shift operator >>> ; this means that you are not wearing a sign bit.

As for (1L << nrBits) - 1L , i.e. 2^nrBits - 1 . L must have long constants.

Also note that there is no “boundary check” (for example, offset or number of bits greater than 63 or negative).

Answer 2

To extract the bits between bit x and bit y, where x is the larger of the two numbers, you could say

 long mask = (Math.pow(2,x+1)-1)- (Math.pow(2,y+1)-1); long extract = input & mask;