Java Bit Operation on Long - Counting Set and Unset bits

Question

I have a long number. Now what I want is following (given in pseudo code) ,

int cnt1 = 0 
int cnt2 = 0 

for each two bits of that long

       if the two bi         


        

Answer 1

A short answer.

long num = ~0L;
int cnt1 = Long.bitCount(num & (num >>> 1) & 0x5555555555555555L);
System.out.println(cnt1);
int cnt2 = 32 - cnt1;

prints

32

Answer 2

What you need to do is create a bit-mask and run it over your value, assuming this is homework I will only give some pointers:

• the bit-mask you already gave: long mask = 0x03L;
• to check every other 2 bits, shift your mask left 2 potisions
• you can use a for-loop to check the value until your mask has value 0
• use the bitwise-and operator & to check a value against a mask

If you put above hints into code, you will have your answer :-)

Edit now that the results are in, my solution would be:

long cnt1 = 0;
long cnt2 = 0;

for (long mask = 0x03; mask != 0; mask <<=2) {

    (mask == (value & mask)) ? cnt1++ : cnt2++;
}

Answer 3

What you need to do is keep shifting by 2 bits to the right at every iteration and do a bitwise and (&) operation with the number 3 (11 in binary):

long number;
int cnt1 = 0;
int cnt2 = 0;
long test = 3;
int counter = 0;    

while(counter < 64) { // we have 64 bits to inspect
    if((number & test) == 3) { // last 2 bits are 11
        cnt1++;
    } else { // last 2 bits are either 10, 01 or 00
        cnt2++;
    }          
    counter += 2;
    number = number >>> 2; // shift by 2 bits to the right
}