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) {

}

我不知道从哪里开始编写这个函数，我认为我将值作为数组循环并对它们应用xor操作。会有类似下面的工作吗？如果没有，那么接近这个的方法是什么？

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”的方式迭代这些位，位于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，为O（1位数）：

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，在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

但您可以使用按位操作。

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

假设有32位整数，有一种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;
}

甚至是unsigned int的奇偶校验

2 个答案: