How To Find The Leading Number Of Zero's In a Number using C

Question

for example,if i have number 64,then its binary representation would be 0000 0000 0000 0000 0000 0000 0100 0000 so leading number of zero\'s is 25. remember i have to calculate

Answer 1

I would go with:

unsigned long clz(unsigned long n) {
    unsigned long result = 0;
    unsigned long mask = 0;
    mask = ~mask;
    auto size = sizeof(n) * 8;
    auto shift = size / 2;
    mask >>= shift;
    while (shift >= 1) {
        if (n <= mask) {
            result += shift;
            n <<= shift;
        }
        shift /= 2;
        mask <<= shift;
    }
    return result;
}

Answer 2

Right shift is your friend.

    int input = 64;
    int sample = ( input < 0 ) ? 0 : input;
    int leadingZeros = ( input < 0 ) ? 0 : 32;

    while(sample) {
        sample >>= 1;
        --leadingZeros;
    }
    printf("Input = %d, leading zeroes = %d\n",input, leadingZeros);

Answer 3

I just found this problem at the top of the search results and this code:

int pop(unsigned x) {
    unsigned n;
    n = (x >> 1) & 033333333333;
    x = x - n;
    n = (n >> 1) & 033333333333;
    x = x - n;
    x = (x + (x >> 3)) & 030707070707;
    return x % 63;
}

int nlz(unsigned x) {
    x = x | (x >> 1);
    x = x | (x >> 2);
    x = x | (x >> 4);
    x = x | (x >> 8);
    x = x | (x >>16);
    return pop(~x);
}

where pop counts 1 bits, is several times faster than the first (upvoted) answer.

I didn't notice, question was about 64 bits numbers, so here:

int nlz(unsigned long x) {
    unsigned long y;
    long n, c;
    n = 64;
    c = 32;
    do {
        y = x >> c;
        if (y != 0) {
            n = n - c;
            x = y;
        }
        c = c >> 1;
    } while (c != 0);
    return n - x;
}

is a 64 bits algorithm, again several times faster than the mentioned above.

Answer 4

See here for the 32-bit version and other great bit-twiddling hacks.

// this is like doing a sign-extension
// if original value was   0x00.01yyy..y
// then afterwards will be 0x00.01111111
x |= (x >> 1);
x |= (x >> 2);
x |= (x >> 4);
x |= (x >> 8);
x |= (x >> 16);
x |= (x >> 32);

and after that you just need to return 64 - numOnes(x). A simple way to do that is numOnes32(x) + numOnes32(x >> 32) where numOnes32 is defined as:

int numOnes32(unsigned int x) {
    x -= ((x >> 1) & 0x55555555);
    x = (((x >> 2) & 0x33333333) + (x & 0x33333333));
    x = (((x >> 4) + x) & 0x0f0f0f0f);
    x += (x >> 8);
    x += (x >> 16);
    return(x & 0x0000003f);
}

I haven't tried out this code, but this should do numOnes64 directly (in less time):

int numOnes64(unsigned long int x) {
     x = ((x >> 1) & 0x5555555555555555L) + (x & 0x5555555555555555L);
     x = ((x >> 2) & 0x3333333333333333L) + (x & 0x3333333333333333L);
     // collapse:
     unsigned int v = (unsigned int) ((x >>> 32) + x);
     v = ((v >> 4) + v) & 0x0f0f0f0f) + (v & 0x0f0f0f0f);
     v = ((v >> 8) & 0x00ff00ff) + (v & 0x00ff00ff);
     return ((v >> 16) & 0x0000ffff) + (v & 0x0000ffff);
}

Answer 5

Because the logarithm base 2 roughly represents the number of bits required to represent a number, it might be useful in the answer:

irb(main):012:0> 31 - (Math::log(64) / Math::log(2)).floor()
=> 25
irb(main):013:0> 31 - (Math::log(65) / Math::log(2)).floor()
=> 25
irb(main):014:0> 31 - (Math::log(127) / Math::log(2)).floor()
=> 25
irb(main):015:0> 31 - (Math::log(128) / Math::log(2)).floor()
=> 24

Of course, one downside to using log(3) is that it is a floating-point routine; there are probably some supremely clever bit-tricks to find the number of leading zero bits in integers, but I can't think of one off the top of my head...

Answer 6

Using floating points is not the right answer....

Here is an algo that I use to count the TRAILING 0... change it for Leading... This algo is in O(1) (will always execute in ~ the same time, or even the same time on some CPU).

int clz(unsigned int i)
{
  int zeros;
  if ((i&0xffff)==0) zeros= 16, i>>= 16; else zeroes= 0;
  if ((i&0xff)==0) zeros+= 8, i>>= 8;
  if ((i&0xf)==0) zeros+= 4, i>>= 4;
  if ((i&0x3)==0) zeros+= 2, i>>= 2;
  if ((i&0x1)==0) zeros+= 1, i>>= 1;
  return zeroes+i;
}