c - 无符号整型的偶校验

Question

/*A value has even parity if it has an even number of 1 bits.
 *A value has an odd parity if it has an odd number of 1 bits.
 *For example, 0110 has even parity, and 1110 has odd parity.
 *Return 1 iff x has even parity.
 */

int has_even_parity(unsigned int x) {

}

我不确定从哪里开始编写这个函数，我想我将值作为数组循环并对它们应用异或运算。 会像下面这样工作吗？如果没有，有什么方法可以解决这个问题？

int has_even_parity(unsigned int x) {
    int i, result = x[0];
    for (i = 0; i < 3; i++){
        result = result ^ x[i + 1];
    }
    if (result == 0){
        return 1;
    }
    else{
        return 0;
    }
}

Answer 1

选项 #1 - 以“显而易见”的方式迭代位，时间复杂度为 O(位数):

int has_even_parity(unsigned int x)
{
    int p = 1;
    while (x)
    {
        p ^= x&1;
        x >>= 1; // at each iteration, we shift the input one bit to the right
    }
    return p;

选项 #2 - 仅迭代设置为 1 的位，时间复杂度为 O(1 的数量):

int has_even_parity(unsigned int x)
{
    int p = 1;
    while (x)
    {
        p ^= 1;
        x &= x-1; // at each iteration, we set the least significant 1 to 0
    }
    return p;
}

选项 #3 - 使用 SWAR 算法以 O(log(位数)) 计算 1:

http://aggregate.org/MAGIC/#Population%20Count%20%28Ones%20Count%29