C ++ next float with numeric_limits / epsilon？ [英] C++ next float with numeric_limits / epsilon?

问题描述

在C ++中考虑一个正常实数 TREAL x （不是subnormal，不是NaN /无限）（ TREAL = float ， double ， long double ） >
以下是从浮点角度查找上一个和下一个 x 的好解决方案？

Consider a "normal" real number TREAL x in C++ (not subnormal and not NaN/Infinite) (TREAL = float, double, long double)
Is the following the good solution to find the previous and next x from a floating-point point of view ?

TREAL xprev = (((TREAL)(1.)) - std::numeric_limits<TREAL>::epsilon()) * x;
TREAL xnext = (((TREAL)(1.)) + std::numeric_limits<TREAL>::epsilon()) * x;

非常感谢。

推荐答案

C99和C ++ 11在< math.h> 和中有nextafter，nextafterl和nextafterf函数; cmath> 。使用基本算术和epsilon实现它们将是乏味的，因为您需要考虑四舍五入。使用二进制表示法可能更容易，但我不知道符号和幅度表示的效果和-0.0的存在（请参阅 Fred's回答需要什么）。

C99 and C++11 have nextafter, nextafterl and nextafterf functions in <math.h> and <cmath>. Implementing them with basic arithmetic and epsilon would be tedious as you'd need to take rounding into account. Working on the binary representation is probably easier, but I wonder about the effect of the sign and magnitude representation and the existence of -0.0 (see Fred's answer for what is needed).

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