如何实现快速逆sqrt没有未定义的行为？ [英] How to implement fast inverse sqrt without undefined behavior?

问题描述

根据我对严格别名规则的了解，此代码< a href =http://en.wikipedia.org/wiki/Fast_inverse_square_root>快速反向平方根将导致C ++中未定义的行为：

  float Q_rsqrt（float number）
 {
 long i; 
 float x2，y; 
 const float threehalfs = 1.5F; 
 
 x2 = number * 0.5F; 
 y = number; 
 i = *（long *）& y; // type punning 
 i = 0x5f3759df  - （i>> 1）; 
 y = *（float *）& i; 
 y = y *（threehalfs  - （x2 * y * y））; 
 
 return y; 
} 
  

这段代码是否确实导致UB？如果是，如何以标准兼容的方式重新实现？

假设：在调用此函数之前，我们以某种方式检查了浮点数是否为IEEE 754 32位格式， code> sizeof（long）== sizeof（float）并且平台是little-endian。

解决方案

您应该使用 memcpy 。 AFAIK这是唯一符合标准的方式，编译器足够聪明，可以用单个字移动指令替换调用。有关这些声明的说明，请参阅此问题。

From what I understood about strict aliasing rule, this code for fast inverse square root will result in undefined behavior in C++:

float Q_rsqrt( float number )
{
    long i;
    float x2, y;
    const float threehalfs = 1.5F;

    x2 = number * 0.5F;
    y  = number;
    i  = * ( long * ) &y; // type punning
    i  = 0x5f3759df - ( i >> 1 );
    y  = * ( float * ) &i;
    y  = y * ( threehalfs - ( x2 * y * y ) );

    return y;
}

Does this code indeed cause UB? If yes, how can it be reimplemented in standard compliant way? If not, why not?

Assumptions: before calling this function we have somehow checked that floats are in IEEE 754 32-bit format, sizeof(long)==sizeof(float) and the platform is little-endian.

解决方案

You should use memcpy. AFAIK this is the only standard compliant way and compilers are smart enough to replace the call with a single word move instruction. See this question for reasoning behind these claims.

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