一个数二进制中1的个数(Hamming weight)

```//types and constants used in the functions below
//uint64_t is an unsigned 64-bit integer variable type (defined in C99 version of C language)
const uint64_t m1  = 0x5555555555555555; //binary: 0101...
const uint64_t m2  = 0x3333333333333333; //binary: 00110011..
const uint64_t m4  = 0x0f0f0f0f0f0f0f0f; //binary:  4 zeros,  4 ones ...
const uint64_t m8  = 0x00ff00ff00ff00ff; //binary:  8 zeros,  8 ones ...
const uint64_t m16 = 0x0000ffff0000ffff; //binary: 16 zeros, 16 ones ...
const uint64_t m32 = 0x00000000ffffffff; //binary: 32 zeros, 32 ones
const uint64_t hff = 0xffffffffffffffff; //binary: all ones
const uint64_t 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.
//This algorithm uses 24 arithmetic operations (shift, add, and).
int popcount64a(uint64_t 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;
}
```

ab – 0a 得到的值为ab中1的个数

```//This uses fewer arithmetic operations than any other known
//implementation on machines with slow multiplication.
//This algorithm uses 17 arithmetic operations.
int popcount64b(uint64_t 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 & 0x7f;
}

//This uses fewer arithmetic operations than any other known
//implementation on machines with fast multiplication.
//This algorithm uses 12 arithmetic operations, one of which is a multiply.
int popcount64c(uint64_t 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) + ...
}
```