寻求无符号数右移向偶舍入函数的简洁等价实现

背景与问题 #

对无符号数执行带舍入的右移操作时，已有以下可行实现：

// 向下舍入右移
unsigned rshift_round_down(unsigned x, unsigned a) {
  return x >> a;
}

// 向上舍入右移
unsigned rshift_round_up(unsigned x, unsigned a) {
  return (x >> a) + ((x & ((1U << a) - 1)) != 0);
}

// 中间值向上舍入右移
unsigned rshift_round_halfway(unsigned x, unsigned a) {
  if (a == 0) return x;
  return (x >> a) + ((x >> (a - 1)) & 1);
}

// 向偶数舍入右移（原实现）
unsigned rshift_round_towards_even(unsigned x, unsigned a) {
  if (a == 0) return x;
  return (x >> a) + ((x >> (a - ((x & ((1U << a) - 1)) != 1 << (a - 1)))) & 1);
}

现询问：是否存在更简洁但功能完全等价的rshift_round_towards_even实现？

注：参数a的取值范围为 0 <= a && a < sizeof(unsigned) * 8，无需支持超出该范围的取值。

示例输入输出 #

• rshift_round_towards_even(60, 2) == 15
• rshift_round_towards_even(61, 2) == 15
• rshift_round_towards_even(62, 2) == 16 （中间值向偶数舍入，结果向上）
• rshift_round_towards_even(63, 2) == 16
• rshift_round_towards_even(64, 2) == 16
• rshift_round_towards_even(65, 2) == 16
• rshift_round_towards_even(66, 2) == 16 （中间值向偶数舍入，结果向下）
• rshift_round_towards_even(67, 2) == 17
• rshift_round_towards_even(68, 2) == 17
• rshift_round_towards_even(69, 2) == 17
• rshift_round_towards_even(70, 2) == 18 （中间值向偶数舍入，结果向上）
• rshift_round_towards_even(71, 2) == 18
• rshift_round_towards_even(72, 2) == 18
• rshift_round_towards_even(73, 2) == 18

简洁等价实现 #

有，基于向偶数舍入的核心规则：

1. 若舍去部分大于2^(a-1)，则进位；
2. 若舍去部分小于2^(a-1)，则不进位；
3. 若舍去部分恰好等于2^(a-1)（中间值），仅当右移后的结果为奇数时进位，否则不进位。

易读版实现

unsigned rshift_round_towards_even(unsigned x, unsigned a) {
  if (a == 0) return x;
  const unsigned mask = (1U << a) - 1;
  const unsigned lower_bits = x & mask; // 被舍去的低a位
  const unsigned upper_bits = x >> a;  // 右移后的高位部分
  // 判断是否需要进位
  const int need_round = (lower_bits > (1U << (a-1))) || 
                         (lower_bits == (1U << (a-1)) && (upper_bits & 1));
  return upper_bits + need_round;
}

紧凑位运算版

如果追求代码紧凑，可将逻辑合并为一行（可读性稍降，但功能完全等价）：

unsigned rshift_round_towards_even(unsigned x, unsigned a) {
  return a ? (x >> a) + (((x >> (a-1)) & 1) & ((x & ((1U << a)-1)) != (1U << (a-1)) || ((x >> a) & 1))) : x;
}

高效运算版

利用无符号数加法特性，还能写出更高效的实现：

unsigned rshift_round_towards_even(unsigned x, unsigned a) {
  if (a == 0) return x;
  const unsigned half = 1U << (a-1);
  const unsigned mask = (1U << a) - 1;
  // 先按中间值向上舍入，再修正中间值时的偶数舍入规则
  return (x + half - 1) >> a - ((x & mask) == half && !((x >> a) & 1));
}

以上实现均与原函数功能完全一致，但代码更简洁、可读性更强。

内容的提问来源于stack exchange，提问作者pts