Count number of bits in an unsigned integer

问题

I want to write a function named bitCount() in the file: bitcount.c that returns the number of bits in the binary representation of its unsigned integer argument.

Here is what I have so far:

#include <stdio.h>

int bitCount (unsigned int n);

int main () {
    printf ("# 1-bits in base 2 representation of %u = %d, should be 0\n",
        0, bitCount (0));
    printf ("# 1-bits in base 2 representation of %u = %d, should be 1\n",
        1, bitCount (1));
    printf ("# 1-bits in base 2 representation of %u = %d, should be 16\n",
        2863311530u, bitCount (2863311530u));
    printf ("# 1-bits in base 2 representation of %u = %d, should be 1\n",
        536870912, bitCount (536870912));
    printf ("# 1-bits in base 2 representation of %u = %d, should be 32\n",
        4294967295u, bitCount (4294967295u));
    return 0;
}

int bitCount (unsigned int n) {
    /* your code here */
}

Okay, when I just run this I get:

# 1-bits in base 2 representation of 0 = 1, should be 0
# 1-bits in base 2 representation of 1 = 56, should be 1
# 1-bits in base 2 representation of 2863311530 = 57, should be 16
# 1-bits in base 2 representation of 536870912 = 67, should be 1
# 1-bits in base 2 representation of 4294967295 = 65, should be 32

RUN SUCCESSFUL (total time: 14ms)

It doesn't return the correct numbers of bits.

What's the best way to return the number of bits in the binary representation of its unsigned integer argument in C?

回答1:

 int bitCount(unsigned int n) {

    int counter = 0;
    while(n) {
        counter += n % 2;
        n >>= 1;
    }
    return counter;
 }

回答2:

Here's a solution that doesn't need to iterate. It takes advantage of the fact that adding bits in binary is completely independent of the position of the bit and the sum is never more than 2 bits. 00+00=00, 00+01=01, 01+00=01, 01+01=10. The first addition adds 16 different 1-bit values simultaneously, the second adds 8 2-bit values, and each one after does half as many until there's only one value left.

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

This is hard-coded to 32 bit integers, if yours are a different size it will need adjusting.

回答3:

Turns out there are some pretty sophisticated ways to compute this as answered here.

The following impl (I learned way back) simply loops knocking off the least significant bit on each iteration.

int bitCount(unsigned int n) {

  int counter = 0;
  while(n) {
    counter ++;
    n &= (n - 1);
  }
  return counter;
}

来源：https://stackoverflow.com/questions/14780928/count-number-of-bits-in-an-unsigned-integer

标签

binary

bit-manipulation