Even parity of a unsigned int [duplicate]

/*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) {

}

I'm not sure where to begin writing this function, I'm thinking that I loop through the value as an array and apply xor operations on them. Would something like the following work? If not, what is the way to approach this?

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;
    }
}

Option #1 - iterate the bits in the "obvious" way, at O(number of bits):

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;

Option #2 - iterate only the bits that are set to 1, at O(number of 1s):

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;
}

Option #3 - use the SWAR algorithm for counting 1s, at O(log(number of bits)):

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

You can't access an integer as an array,

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

But you can use bitwise operations.

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

There is a clever way to do this, assuming 32-bit integers:

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

来源：https://stackoverflow.com/questions/21589674/even-parity-of-a-unsigned-int

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bits

parity