c++ - 如何在 C++ 中实现 Fortran 间距()函数？

Question

我从 Fortran 转换为 C++ 的代码包含 spacing(x)功能。从描述来看，spacing(x)返回

Smallest distance between two numbers of a given type

和

Determines the distance between the argument X and the nearest adjacent number of the same type.

是否有 C++ 等效函数，或者如果没有，我如何在 C++ 中实现该函数？

Answer 1

使用 SPACING如 Determines the distance between the argument X and the nearest adjacent number of the same type , 使用 nexttoward() .

upper = nexttoward(x, INFINITY) - x;
lower = x - nexttoward(x, -INFINITY);
spacing = fmin(upper, lower);

 

upper != lower在特定情况下:例如 x是 2 的幂。
可能需要一些工作来处理缺乏真正 INFINITY 的实现。
或

if (x > 0) {
  spacing = x - nexttoward(x, 0);
} else {
  // 1.0 used here instead of 0 to handle x==0
  spacing = nexttoward(x, 1.0) - x; 
}

或

// Subtract next smaller-in-magnitude value.  With 0, use next toward 1.
spacing = fabs(x - nexttoward(x, !x));

我怀疑 nextafter()效果与 nexttoward() 一样好，甚至更好.