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# Counting the Number of Bits in a Byte

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Counting Number of On Bits in an Integer

## Fast Bit Counting Routines

Compiled from various sources by Gurmeet Singh Manku

A common problem asked in job interviews is to count the number of bits that are on in an unsigned integer. Here are seven solutions to this problem. Source code in C is available.

 Iterated Count ``` int bitcount (unsigned int n) { int count=0; while (n) { count += n & 0x1u ; n >>= 1 ; } return count ; } ```
 Sparse Ones ``` int bitcount (unsigned int n) { int count=0 ; while (n) { count++ ; n &= (n - 1) ; } return count ; } ```
 Dense Ones ``` int bitcount (unsigned int n) { int count = 8 * sizeof(int) ; n ^= (unsigned int) -1 ; while (n) { count-- ; n &= (n - 1) ; } return count ; } ```
 Precompute_8bit ``` // static int bits_in_char  ; int bitcount (unsigned int n) { // works only for 32-bit ints return bits_in_char [n & 0xffu] + bits_in_char [(n >> 8) & 0xffu] + bits_in_char [(n >> 16) & 0xffu] + bits_in_char [(n >> 24) & 0xffu] ; } ```
Iterated Count runs in time proportional to the total number of bits. It simply loops through all the bits, terminating slightly earlier because of the while condition. Useful if 1's are sparse and among the least significant bits. Sparse Ones runs in time proportional to the number of 1 bits. The line n &= (n - 1) simply sets the rightmost 1 bit in n to 0. Dense Ones runs in time proportional to the number of 0 bits. It is the same as Sparse Ones, except that it first toggles all bits (n ~= -1), and continually subtracts the number of 1 bits from sizeof(int). Precompute_8bit assumes an array bits_in_char such that bits_in_char[i] contains the number of 1 bits in the binary representation for i. It repeatedly updates count by masking out the last eight bits in n, and indexing into bits_in_char.

 Precompute_16bit ``` // static char bits_in_16bits [0x1u > 16) & 0xffffu] ; } ```
Precompute_16bit is a variant of Precompute_8bit in that an array bits_in_16bits[] stores the number of 1 bits in successive 16 bit numbers (shorts).

 Parallel Count ``` #define TWO(c) (0x1u > (TWO(c))) & MASK(c)) int bitcount (unsigned int n) { n = COUNT(n, 0) ; n = COUNT(n, 1) ; n = COUNT(n, 2) ; n = COUNT(n, 3) ; n = COUNT(n, 4) ; /* n = COUNT(n, 5) ; for 64-bit integers */ return n ; } ```
Parallel Count carries out bit counting in a parallel fashion. Consider n after the first line has finished executing. Imagine splitting n into pairs of bits. Each pair contains the number of ones in those two bit positions in the original n. After the second line has finished executing, each nibble contains the number of ones in those four bits positions in the original n. Continuing this for five iterations, the 64 bits contain the number of ones among these sixty-four bit positions in the original n. That is what we wanted to compute.

 Nifty Parallel Count ``` #define MASK_01010101 (((unsigned int)(-1))/3) #define MASK_00110011 (((unsigned int)(-1))/5) #define MASK_00001111 (((unsigned int)(-1))/17) int bitcount (unsigned int n) { n = (n & MASK_01010101) + ((n >> 1) & MASK_01010101) ; n = (n & MASK_00110011) + ((n >> 2) & MASK_00110011) ; n = (n & MASK_00001111) + ((n >> 4) & MASK_00001111) ; return n % 255 ; } ```
Nifty Parallel Count works the same way as Parallel Count for the first three iterations. At the end of the third line (just before the return), each byte of n contains the number of ones in those eight bit positions in the original n. A little thought then explains why the remainder modulo 255 works.

 MIT HACKMEM Count ``` int bitcount(unsigned int n) { /* works for 32-bit numbers only */ /* fix last line for 64-bit numbers */ register unsigned int tmp; tmp = n - ((n >> 1) & 033333333333) - ((n >> 2) & 011111111111); return ((tmp + (tmp >> 3)) & 030707070707) % 63; } ```
MIT Hackmem Count is funky. Consider a 3 bit number as being 4a+2b+c. If we shift it right 1 bit, we have 2a+b. Subtracting this from the original gives 2a+b+c. If we shift the original 2 bits right we get a, and so with another subtraction we have a+b+c, which is the number of bits in the original number. How is this insight employed? The first assignment statement in the routine computes tmp. Consider the octal representation of tmp. Each digit in the representation is simply the number of 1's in the corresponding three bit positions in n. The last return statement sums these octal digits to produce the final answer. The key idea is to add adjacent pairs of octal digits together and then compute the remainder modulus 63. This is accomplished by right-shifting tmp by three bits, adding it to tmp itself and ANDing with a suitable mask. This yields a number in which groups of six adjacent bits (starting from the LSB) contain the number of 1's among those six positions in n. This number modulo 63 yields the final answer. For 64-bit numbers, we would have to add triples of octal digits and use modulus 1023. This is HACKMEM 169, as used in X11 sources. Source: MIT AI Lab memo, late 1970's.

 ``` No Optimization Some Optimization Heavy Optimization Precomp_16 52.94 Mcps Precomp_16 76.22 Mcps Precomp_16 80.58 Mcps Precomp_8 29.74 Mcps Precomp_8 49.83 Mcps Precomp_8 51.65 Mcps Parallel 19.30 Mcps Parallel 36.00 Mcps Parallel 38.55 Mcps MIT 16.93 Mcps MIT 17.10 Mcps Nifty 31.82 Mcps Nifty 12.78 Mcps Nifty 16.07 Mcps MIT 29.71 Mcps Sparse 5.70 Mcps Sparse 15.01 Mcps Sparse 14.62 Mcps Dense 5.30 Mcps Dense 14.11 Mcps Dense 14.56 Mcps Iterated 3.60 Mcps Iterated 3.84 Mcps Iterated 9.24 Mcps Mcps = Million counts per second ```
Which of the several bit counting routines is the fastest? Results of speed trials on an i686 are summarized in the table on left. "No Optimization" was compiled with plain gcc. "Some Optimizations" was gcc -O3. "Heavy Optimizations" corresponds to gcc -O3 -mcpu=i686 -march=i686 -fforce-addr -funroll-loops -frerun-cse-after-loop -frerun-loop-opt -malign-functions=4.
Gurmeet Singh Manku
Last update: 27 Jul 2002