汉明权重 - 维基百科，自由的百科全书

//types and constants used in the functions below  typedef unsigned __int64 uint64;  //assume this gives 64-bits const uint64 m1 = 0x5555555555555555; //binary: 0101... const uint64 m2 = 0x3333333333333333; //binary: 00110011.. const uint64 m4 = 0x0f0f0f0f0f0f0f0f; //binary:  4 zeros,  4 ones ... const uint64 m8 = 0x00ff00ff00ff00ff; //binary:  8 zeros,  8 ones ... const uint64 m16 = 0x0000ffff0000ffff; //binary: 16 zeros, 16 ones ... const uint64 m32 = 0x00000000ffffffff; //binary: 32 zeros, 32 ones ... const uint64 hff = 0xffffffffffffffff; //binary: all ones const uint64 h01 = 0x0101010101010101; //the sum of 256 to the power of 0,1,2,3...  //This is a naive implementation, shown for comparison, //and to help in understanding the better functions. //It uses 24 arithmetic operations (shift, add, and). int popcount_1(uint64 x) {     x = (x & m1 ) + ((x >>  1) & m1 ); //put count of each  2 bits into those  2 bits      x = (x & m2 ) + ((x >>  2) & m2 ); //put count of each  4 bits into those  4 bits      x = (x & m4 ) + ((x >>  4) & m4 ); //put count of each  8 bits into those  8 bits      x = (x & m8 ) + ((x >>  8) & m8 ); //put count of each 16 bits into those 16 bits      x = (x & m16) + ((x >> 16) & m16); //put count of each 32 bits into those 32 bits      x = (x & m32) + ((x >> 32) & m32); //put count of each 64 bits into those 64 bits      return x; }  //This uses fewer arithmetic operations than any other known   //implementation on machines with slow multiplication. //It uses 17 arithmetic operations. int popcount_2(uint64 x) {     x -= (x >> 1) & m1;             //put count of each 2 bits into those 2 bits     x = (x & m2) + ((x >> 2) & m2); //put count of each 4 bits into those 4 bits      x = (x + (x >> 4)) & m4;        //put count of each 8 bits into those 8 bits      x += x >>  8;  //put count of each 16 bits into their lowest 8 bits     x += x >> 16;  //put count of each 32 bits into their lowest 8 bits     x += x >> 32;  //put count of each 64 bits into their lowest 8 bits     return x &0xff; }  //This uses fewer arithmetic operations than any other known   //implementation on machines with fast multiplication. //It uses 12 arithmetic operations, one of which is a multiply. int popcount_3(uint64 x) {     x -= (x >> 1) & m1;             //put count of each 2 bits into those 2 bits     x = (x & m2) + ((x >> 2) & m2); //put count of each 4 bits into those 4 bits      x = (x + (x >> 4)) & m4;        //put count of each 8 bits into those 8 bits      return (x * h01)>>56;  //returns left 8 bits of x + (x<<8) + (x<<16) + (x<<24) + ...  } 

//This is better when most bits in x are 0 //It uses 3 arithmetic operations and one comparison/branch per "1" bit in x. int popcount_4(uint64 x) {     uint64 count;     for (count=0; x; count++)         x &= x-1;     return count; }  //This is better if most bits in x are 0. //It uses 2 arithmetic operations and one comparison/branch  per "1" bit in x. //It is the same as the previous function, but with the loop unrolled. #define f(y) if ((x &= x-1) == 0) return y; int popcount_5(uint64 x) {     if (x == 0) return 0;     f( 1) f( 2) f( 3) f( 4) f( 5) f( 6) f( 7) f( 8)     f( 9) f(10) f(11) f(12) f(13) f(14) f(15) f(16)     f(17) f(18) f(19) f(20) f(21) f(22) f(23) f(24)     f(25) f(26) f(27) f(28) f(29) f(30) f(31) f(32)     f(33) f(34) f(35) f(36) f(37) f(38) f(39) f(40)     f(41) f(42) f(43) f(44) f(45) f(46) f(47) f(48)     f(49) f(50) f(51) f(52) f(53) f(54) f(55) f(56)     f(57) f(58) f(59) f(60) f(61) f(62) f(63)     return 64; }  //Use this instead if most bits in x are 1 instead of 0 #define f(y) if ((x |= x+1) == hff) return 64-y;