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

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++;
}