@无名啊，这是这个函数的PHP版本，有助于理解为什么没有未定义行为：
<?php function Q_rsqrt(float $number) { $threehalfs = 1.5; $x2 = $number * 0.5; $y = $number; $i = unpack("l", pack("f", $y))[1]; $i = 0x5f3759df - ($i >> 1); $y = unpack("f", pack("l", $i))[1]; $y = $y * ( $threehalfs - ($x2 * $y * $y) ); $y = $y * ( $threehalfs - ($x2 * $y * $y) ); return $y; } printf("%0.7f\n", Q_rsqrt(3.14)); printf("%0.7f\n", Q_rsqrt(1024.0)); printf("%0.7f\n", Q_rsqrt(10086.0)); printf("%0.7f\n", Q_rsqrt(2147483647.0)); printf("%0.14f\n", Q_rsqrt(3.14)); printf("%0.14f\n", Q_rsqrt(1024.0)); printf("%0.14f\n", Q_rsqrt(10086.0)); printf("%0.14f\n", Q_rsqrt(2147483647.0));在给定的定义域和有效数字范围内，它和C版本的结果一致。如果继续增加输出的位数，结果就开始不一致了，因为PHP在内部使用64位整数和双精度浮点数，而非C代码的32位整数和单精度浮点数，只在
pack和unpack时才转换为32位单精度，所以两者会有精度差异。此外32位和64位在处理符号位上可能也有差异，所以C版给出负数解的情况下PHP给出的是正数解。当然两者都是正确的解，因为负数的平方也是正数。
C版本：
#include <stdio.h> float Q_rsqrt( float number ) { long i; float x2, y; const float threehalfs = 1.5F; x2 = number * 0.5F; y = number; i = * ( long * ) &y; // evil floating point bit level hacking i = 0x5f3759df - ( i >> 1 ); // what the fuck? y = * ( float * ) &i; y = y * ( threehalfs - ( x2 * y * y ) ); // 1st iteration y = y * ( threehalfs - ( x2 * y * y ) ); // 2nd iteration, this can be removed return y; } int main() { printf("%0.7f\n", Q_rsqrt(3.14)); printf("%0.7f\n", Q_rsqrt(1024.0)); printf("%0.7f\n", Q_rsqrt(10086.0)); printf("%0.7f\n", Q_rsqrt(2147483647.0)); printf("%0.14f\n", Q_rsqrt(3.14)); printf("%0.14f\n", Q_rsqrt(1024.0)); printf("%0.14f\n", Q_rsqrt(10086.0)); printf("%0.14f\n", Q_rsqrt(2147483647.0)); return 0; }
能用纯SQL实现
要我的话肯定得用常规编程语言，甚至上人工智能
老虎会游泳
