Haskell 范围和 float

Question

为什么 Haskell 范围表示法对于 float 的行为与整数和字符不同？

Prelude> [1, 3 .. 10] :: [Int]
[1,3,5,7,9] 
Prelude> [1, 3 .. 10] :: [Float]
[1.0,3.0,5.0,7.0,9.0,11.0]
Prelude> ['a', 'c' .. 'f']
"ace"

如果最后一个元素接近上限，我会理解，但这显然不是舍入问题。

Answer 1

语法[e1, e2 .. e3]确实是 enumFromThenTo e1 e2 e3 的语法糖，这是 Enum 中的函数类型类。

The Haskell standard其语义定义如下:

For the types Int and Integer, the enumeration functions have the following meaning:

• The sequence enumFrom e1 is the list [e1,e1 + 1,e1 + 2,…].
• The sequence enumFromThen e1 e2 is the list [e1,e1 + i,e1 + 2i,…], where the increment, i, is e2 − e1. The increment may be zero or negative. If the increment is zero, all the list elements are the same.
• The sequence enumFromTo e1 e3 is the list [e1,e1 + 1,e1 + 2,…e3]. The list is empty if e1 > e3.
• The sequence enumFromThenTo e1 e2 e3 is the list [e1,e1 + i,e1 +
2i,…e3], where the increment, i, is e2 − e1. If the increment is positive or zero, the list terminates when the next element would be greater than e3; the list is empty if e1 > e3. If the increment is negative, the list terminates when the next element would be less than e3; the list is empty if e1 < e3.

这几乎是您所期望的，但是 Float和Double实例的定义不同:

For Float and Double, the semantics of the enumFrom family is given by the rules for Int above, except that the list terminates when the elements become greater than e3 + i∕2 for positive increment i, or when they become less than e3 + i∕2 for negative i.

我不太确定这样做的理由是什么，所以我能给你的唯一答案是它就是这样，因为它在标准中是这样定义的。

您可以通过使用整数进行枚举并转换为 Float 来解决此问题。之后。

Prelude> map fromIntegral [1, 3 .. 10] :: [Float]
[1.0,3.0,5.0,7.0,9.0]