CODEWITHSUNDEEP
javabit-manipulation
Arpssss
Arpssss

Reputation: 3858

Java Long - Bit's manipulation

I have a long number and I want to manipulate it bits in following way:

Long l = "11000000011" (long l 's bit representation)

Long ll1 = "110000000" (remove last two bits from l and convert to Long)

Long ll2 = "11" (keep last two bit's of l and discard other bits and convert to Long)

Can anybody help me, how to do this in Java in a fast way ?

Upvotes: 0

Views: 2314

Answers (2)

Arpssss
Arpssss

Reputation: 3858

 long l = Long.parseLong("11110", 2) ;
 long ll1 = l >>> 2;
 long lb = (l & 1) ; last bit
 long ls = l >>> 1;
 long lb1 = (ls & 1) ; bit before last bit

Upvotes: 0

templatetypedef
templatetypedef

Reputation: 372694

To convert a string of bits into a long, you can use Long.parseLong:

long l = Long.parseLong("11000000011", 2);

You can then use the bit-shifting operators >>, <<, and >>> to drop off the lower bits. For example:

long ll1 = l >>> 2;

To drop off all but the top two bits, you can use Long.bitCount to count the bits, then shift off the remaining bits.

long ll2 = l >>> (Long.bitCount(ll1) - 2);

EDIT: Since the question you're asking has to do with going from longs to the bits of the longs, you should use the Long.toBinaryString method for this:

String bits = Long.toBinaryString(/* value */);

From there, to drop off the last two bits you can use simple string manipulation. Try using String.substring for this.

Hope this helps!

Upvotes: 3

Related Questions