OCaml mod函数与％相比返回不同的结果 [英] OCaml mod function returns different result compared with %

问题描述

与 python 中的模运算符相比， OCaml mod中的模函数返回的结果有所不同.

The modulo function in OCaml mod return results different when compared with the modulo operator in python.

OCaml:

# -1 mod 4
- : int = -1

Python:

>>> -1 % 4
3

为什么结果不同?.

在OCaml中是否有任何标准模块功能可以用作%?

Is there any standard module function that operate as % in OCaml?.

推荐答案

Python在%运算符的用法上有点不同，该运算符实际上计算两个值的 modulo ，而其他运算符编程语言使用相同的运算符计算 remainder .例如，Scheme中的区别很明显:

Python is a bit different in its usage of the % operator, which really computes the modulo of two values, whereas other programming languages compute the remainder with the same operator. For example, the distinction is clear in Scheme:

(modulo -1 4)    ; modulo
=> 3
(remainder -1 4) ; remainder
=> -1

在Python中:

-1 % 4           # modulo
=> 3
math.fmod(-1, 4) # remainder
=> -1

但是根据此当然，您可以根据remainder来实现自己的modulo操作，如下所示:

Of course, you can implement your own modulo operation in terms of remainder, like this:

let modulo x y =
  let result = x mod y in
  if result >= 0 then result
  else result + y

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