如何在 C++ 中实现 Fortran spacing() 函数？

Question

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

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，如果没有，我如何在 C++ 中实现该函数？

Answer 1

将 SPACING 用作 Determines the distance between the argument X and the nearest adjacent number of the same type, use nexttoward()。

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

 

upper != lower 在 select 个案例中：例如x 是 2 的幂。
可能需要一些工作来处理缺少真正无限的实现。

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));

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我怀疑 nextafter() 也能工作，或者比 nexttoward() 更好。