Manuel Manuel - 2 months ago 7
C Question

How to check if value has even parity of bits or odd?

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. Well For example, 0110 has even parity, and 1110 has odd parity. I have to return 1 iff x has even parity.
Got stuck....Any ideas???Any help will appreciated. Thanks. Also if it very simple question the sorry..Very beginner here

int has_even_parity(unsigned int x) {



int has_even_parity(unsigned int x){
    unsigned int count = 0, i, b = 1;

    for(i = 0; i < 32; i++){
        if( x & (b << i) ){count++;}

    if( (count % 2) ){return 0;}

    return 1;