Checking bits in a float

[earth_angel]

Hey,

I'm wandering if there is an easy way to check the number of bits set to 1 in a binary number (4 byte ints). I know I can convert the int into an array of chars (or int for that matter) with each entry representing a bit, but that is time- consuming. I only need to know if there an odd or even numver of 1's in an int.

Any ideas?

Thanks.

[Dave_Sinkula]

http://www.sjbaker.org/steve/software/cute_code.html

[hk_mp5kpdw]

You can also use a bitset if you want:

#include <iostream>
#include <bitset>
#include <limits>
using namespace std;

int main()
{
    unsigned value = 12345;
    bitset<numeric_limits<unsigned>::digits> bits(value);
    
    cout << "Decimal: " << value
         << "\nBinary: " << bits
         << "\n# bits on: " << bits.count()
         << "\n# bits off: " << numeric_limits<unsigned>::digits - bits.count()
         << endl;

    return 0;
}

Output:

Decimal: 12345
Binary: 00000000000000000011000000111001
# bits on: 6
# bits off: 26

[earth_angel]

I can't open Dave's link from the comp I'm using, download limits, etc.

I'l read up a bit more on bitset.

[Dave_Sinkula]

The link had this (among others).

int bits_set(unsigned int n)
{
   n = (n & 0x55555555) + ((n & 0xaaaaaaaa) >> 1);
   n = (n & 0x33333333) + ((n & 0xcccccccc) >> 2);
   n = (n & 0x0f0f0f0f) + ((n & 0xf0f0f0f0) >> 4);
   n = (n & 0x00ff00ff) + ((n & 0xff00ff00) >> 8);
   n = (n & 0x0000ffff) + ((n & 0xffff0000) >> 16);
   return n;
}

[DougDbug]

I'll assume that the title of your post is wrong. It's tricky with a float (which has a mantessa and an exponent)... easy with an int.

You don't need to change the number to a char.

You do need to check one bit at a time... You need a loop that runs 32 times... Count-up the number of "ones" that you find.

You test each bit by performing a bitwise AND on each bit.

X & 1 is true (non-zero) if X bit-zero is 1.
X & 2 is true (non-zero) if X bit-1 is 1.
X & 4 is true (non-zero) if X bit-2 is 1.
X & 8 is true (non-zero) if X bit-3 is 1.
etc...

[Dave_Sinkula]

You need a loop that runs 32 times...

You could loop fewer times, too. Perhaps once for every set bit.

int bits_set(unsigned int n)
{
   int count = 0;
   while ( n )
   {
      n = n & (n - 1); /* clear lsb */
      ++count;
   }
   return count;
}