xoring如何计算比特率?

/*
 * bitParity - returns 1 if x contains an odd number of 0's
 *   Examples: bitParity(5) = 0, bitParity(7) = 1
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 20
 *   Rating: 4
 */
int bitParity(int x) { //XORing two numbers returns a number with same bit parity. 
                       //Keep shifting half of our number until reduced to 1 bit simple case
  x = ( x >> 16 ) ^ x;
  x = ( x >> 8 ) ^ x;
  x = ( x >> 4 ) ^ x;
  x = ( x >> 2 ) ^ x;
  x = ( x >> 1) ^ x;

  return (x & 1);
}

谷歌上提供的大多数解决方案各不相同,

但是在这个比特率问题中,答案是相同的(使用异或)

我知道问这个或那个代码是怎么工作的是错误的,

所以我想知道它的原理和想法。

int bitParity(int x) {
  int x1 = ( x >> 28 ) ^ x;
  int x2 = ( x >> 24 ) ^ x1;
  int x3 = ( x >> 20 ) ^ x2;
  int x4 = ( x >> 16 ) ^ x3;
  int x5 = ( x >> 12 ) ^ x4;
  int x6 = ( x >> 8 ) ^ x5;
  int x7 = ( x >> 4 ) ^ x6;
  int x8 = ( x >> 2 ) ^ x7;
  int x9 = ( x >> 1 ) ^ x8;
  return (x9 & 1);
}

以上代码不起作用