在 Haskell 中返回特定类型 [英] Return specific type within Haskell

问题描述

我有一个关于 Haskell 类型系统的非常普遍的问题.我正在努力熟悉它，我有以下功能:

I have a pretty general question about Haskell's type system. I'm trying to become familiar with it, and I have the following function:

getN :: Num a => a
getN = 5.0 :: Double

当我尝试编译它时，出现以下错误:

When I try to compile this, I get the following error:

Couldn't match expected type `a' against inferred type `Double'
  `a' is a rigid type variable bound by
      the type signature for `getN' at Perlin.hs:15:12
In the expression: 5.0 :: Double
In the definition of `getN': getN = 5.0 :: Double

据我所知，该函数被设置为返回"类 Num 中的一个类型.Double 在这个类中 (http://www.zvon.org/other/haskell/Outputprelude/Num_c.html)，所以我原以为在这种情况下返回"一个 Double 是可以的.

As I understand this, the function is set up to "return" a type in the class Num. Double is in this class (http://www.zvon.org/other/haskell/Outputprelude/Num_c.html), so I would have expected that it would be okay to "return" a Double in this case.

有人可以解释一下吗?

推荐答案

带有签名的函数 Num a =>a 预计适用于 Num 类中的 任何 类型.实现 5.0 :: Double 仅适用于 one 类型，而不适用于该类的 所有 类型，因此编译器会抱怨.

A function with signature Num a => a is expected to work for any type in the class Num. The implementation 5.0 :: Double just works for one type, not for all types of the class, so the compiler complains.

一个通用函数的例子是:

An example of a generic function would be:

square :: (Num a) => a -> a
square x = x * x

这适用于 任何 类型的 Num.它适用于双精度数、整数和您想使用的任何其他数字.因此，它可以有一个泛型类型签名，只需要参数在 Num 类中.(类型类 Num 是必需的，因为该函数使用与 * 的乘法，在那里定义)

This works for any type that is a Num. It works for doubles, integers and whatever other numbers you want to use. Because of that it can have a generic type signature that just requires the parameter to be in class Num. (Type class Num is necessary because the function uses multiplication with *, which is defined there)

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