Number of 1 Bits

求一个无符号32位int有几个1bit。本人只想到了最原始的最末位与0x01与，再右移一位。

解答有一个构造了几个int来按位与的逻辑挺巧妙，只需要与几次就能得出结果，不过短时间内可能构造不出需要的int。本人肯定是不行。

本人代码如下：

int hammingWeight(uint32_t n) {    int result = 0;    while (n > 0) {        if (n & 0x01)            result++;        n = n >> 1;        // n /= 2;    }    // result = popcount_1(n);    return result;}

解答代码如下，且有3种方法：

//types and constants used in the functions below 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 onesconst uint64_t hff = 0xffffffffffffffff; //binary: all onesconst 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.//It uses 24 arithmetic operations (shift, add, and).int popcount_1(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;} //This uses fewer arithmetic operations than any other known  //implementation on machines with slow multiplication.//It uses 17 arithmetic operations.int popcount_2(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.//It uses 12 arithmetic operations, one of which is a multiply.int popcount_3(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) + ... }