c - 无符号整数的偶校验

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其应用异或操作。会像以下工作吗？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“明显”的方式迭代位，在 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 算法计算 1，时间为 O(log(位数))：

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

Answer 2

您不能将整数作为数组访问，

unsigned x = ...;
// x[0]; doesn't work

但是您可以使用按位运算。

unsigned x = ...;
int n = ...;
int bit = (x >> n) & 1u; // Extract bit n, where bit 0 is the LSB

假设是 32 位整数，有一个聪明的方法可以做到这一点：

unsigned parity(unsigned x)
{
    x ^= x >> 16;
    x ^= x >> 8;
    x ^= x >> 4;
    x ^= x >> 2;
    x ^= x >> 1;
    return x & 1;
}