C code to count the number of '1' bits in

问题:

I need C code to return the number of 1's in an unsigned char in C. I need an explanation as to why it works if it's not obvious. I've found a lot of code for a 32-bit number but not much for an unsigned char.

回答1:

The same code will work for an unsigned char. Loop over all bits testing them. See this.

回答2:

const unsigned char oneBits[] = {0,1,1,2,1,2,2,3,1,2,2,3,2,3,3,4};

unsigned char CountOnes(unsigned char x)
{
    unsigned char results;
    results = oneBits[x&0x0f];
    results += oneBits[x>>4];
    return results
}

Have an array that knows the number of bits for 0 through 15. Add the results for each nibble.

回答3:

HACKMEM has this algorithm in 3 operations (roughly translated to C):

bits = (c * 01001001001ULL & 042104210421ULL) % 017;

(ULL is to force 64-bit arithmetic. It's needed, just barely... this calculation requires 33-bit integers.)

Actually, you can replace the second constant with 042104210021ULL, since you're only counting 8 bits, but it doesn't look as nicely symmetrical.

How does this work? Think of c bit-wise, and remember that (a + b) % c = (a % c + b % c) % c, and (a | b) == a + b iff (a & b) == 0.

  (c * 01001001001ULL & 042104210421ULL) % 017
  01   01001001001                01         1
  02   02002002002       02000000000         1
  04   04004004004          04000000         1
 010  010010010010            010000         1
 020  020020020020               020         1
 040  040040040040      040000000000         1  # 040000000000 == 2 ** 32
0100 0100100100100        0100000000         1
0200 0200200200200           0200000         1

If you don't have 64-bit arithmetic available, you can split c up into nibbles and do each half, taking 9 operations. This only requires 13 bits, so using 16- or 32-bit arithmetic will work.

bits = ((c & 017) * 0421 & 0111) % 7 + ((c >> 4) * 0421 & 0111) % 7;

(c * 0421 & 01111) % 7
 1   0421      01    1
 2  01042   01000    1
 4  02104    0100    1
 8  04210     010    1

For example, if c == 105 == 0b11001001,

c == 0100
   |  040
   |  010
   |   01 == 0151
* 01001001001001ULL == 0100100100100
                     |  040040040040
                     |  010010010010
                     |   01001001001 == 0151151151151
& 0421042104210421ULL ==  0100000000
                       | 04000000000
                       |      010000
                       |          01 ==   04100010001
% 017                                == 4

c & 017      ==            8 | 1           ==                   011
011 * 0421   ==     8 * 0421 | 1 * 0421    == 04210 | 0421 == 04631
04631 & 0111 == 04210 & 0111 | 0421 & 0111 ==   010 | 01   ==   011
011 % 7      == 2

c >> 4       ==            4 | 2            ==                     06
06 * 0421    ==     4 * 0421 | 2 * 0421     == 02104 | 01042 == 03146
03146 & 0111 == 02104 & 0111 | 01042 & 0111 ==  0100 | 01000 == 01100
01100 % 7    == 2

2 + 2 == 4

回答4:

See the bit twiddling hacks page: http://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetKernighan

there are many good solutions for this.

Also, this function in its simplest implementation is fairly trivial. You should take the time to learn how to do this.

回答5:

For a integer as small as an unsigned char you get best performance using a small lookup-table.

I know what population-count algorithms you're mentioning. They work by doing arithmetic of multiple words smaller than an integer stored in a register.

This technique is called SWAR (http://en.wikipedia.org/wiki/SWAR).

For more information I suggest you check out the hackers delight website: www.hackersdelight.org. He has example code and written a book that explains these tricks in detail.

回答6:

As already answered, the standard ways of counting bits also work on unsigned chars.

Example:

    unsigned char value = 91;
int bitCount = 0;
while(value > 0)
{
    if ( value & 1 == 1 ) 
        bitCount++;
    value >>= 1;
}

回答7:

an unsigned char is a "number" in just the same way that a 32-bit float or integer is a "number", what the compiler deems them to represent is what changes.

if you picture a char as its bits:

01010011 (8 bits);

you can count the set bits by doing the following:

take the value, lets say x, and take x % 2, the remainder will be either 1 or 0. that is, depending on the endianness of the char, the left or right most bit. accumulate the remainder in a separate variable (this will be the resulting number of set bits).

then >> (right shift) 1 bit.

repeat until 8 bits have been shifted.

the c code should be pretty simple to implement from my pseudocode, but basically

public static int CountSetBits(char c)
{
    int x = 0;
    int setBits = 0;
    while (x < 7)
    {
       setBits = setBits + c % 2;
       c = c >> 1;
       x = x + 1;
    }
}

回答8:

base on Ephemient's post, we have the no branched 8 bits version. It is in hexadecimal expression.

typedef unsigned char       UINT8;
typedef unsigned short      UINT16;
typedef unsigned long long  UINT64;
int hammingWeight8( const UINT8& c)
{
    return ( c* 0x8040201ULL & 0x11111111)%0xF;
}

Apply it twice, we have a 16bits version, which needs 9 operations.

int hammingWeight16( const UINT16& c)
{
    return ((c & 0xFF)* 0x8040201ULL & 0x11111111)%0xF + 
             ((c >> 8)* 0x8040201ULL & 0x11111111)%0xF;
}

Here I write a variant 16bits version which needs 64bits registers and 11 operations. It seems not better than the previous one, but it just uses 1 modulo operation.

int hammingWeight16( const UINT16& c)
{
    UINT64  w;
    w= (((( c* 0x8000400020001ULL)>> 3) & 0x1111111111111111)+14)%0xF;
    return (c!=0)*(w+1+(c==0xFFFF)*15);
}