在Haskell中求幂 [英] Exponentiation in Haskell

问题描述

有人可以告诉我为什么Haskell Prelude为取幂定义了两个单独的函数（即 ^ 和 ** ） ？我认为类型系统应该消除这种重复。

  Prelude> 2 ^ 2 
 4 
 Prelude> 4 ** 0.5 
 2.0 
  

解决方案

实际上有三个指数运算符：（^），（^^）和（**） 。 ^ 是非负整数指数， ^^ 是整数指数， ** 是浮点指数运算：

 （^）::（Num a，Integral b）=> ; a  - > b  - > a 
（^^）::（分数a，积分b）=> a  - > b  - > a 
（**）:: Floating a => a  - > a  - > a 
  

原因是类型安全性：数值运算的结果通常与输入参数具有相同的类型S）。但是，您不能将 Int 提升为浮点数，并得到类型为 Int 的结果。因此，类型系统会阻止您执行此操作：（1 :: Int）** 0.5 会产生类型错误。对于（1 :: Int）^^（-1）。

也是如此。 ： Num 类型在 ^ 下关闭（它们不需要有乘法逆），小于类型在 ^^ 下关闭，浮动类型在<$ c下关闭$ C> ** 。由于 Int 没有小数实例，因此您无法将其提升至负值。

理想情况下， ^ 的第二个参数将被静态约束为非负数（当前 1 ^（-2）引发运行时异常）。但是 Prelude 中没有自然数的类型。

Can someone tell me why the Haskell Prelude defines two separate functions for exponentiation (i.e. ^ and **)? I thought the type system was supposed to eliminate this kind of duplication.

Prelude> 2^2
4
Prelude> 4**0.5
2.0

解决方案

There are actually three exponentiation operators: (^), (^^) and (**). ^ is non-negative integral exponentiation, ^^ is integer exponentiation, and ** is floating-point exponentiation:

(^) :: (Num a, Integral b) => a -> b -> a
(^^) :: (Fractional a, Integral b) => a -> b -> a
(**) :: Floating a => a -> a -> a

The reason is type safety: results of numerical operations generally have the same type as the input argument(s). But you can't raise an Int to a floating-point power and get a result of type Int. And so the type system prevents you from doing this: (1::Int) ** 0.5 produces a type error. The same goes for (1::Int) ^^ (-1).

Another way to put this: Num types are closed under ^ (they are not required to have a multiplicative inverse), Fractional types are closed under ^^, Floating types are closed under **. Since there is no Fractional instance for Int, you can't raise it to a negative power.

Ideally, the second argument of ^ would be statically constrained to be non-negative (currently, 1 ^ (-2) throws a run-time exception). But there is no type for natural numbers in the Prelude.

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