返回Haskell中的特定类型 [英] Return specific type within Haskell

问题描述

我对Haskell的类型系统有一个相当普遍的问题。我试图熟悉它，并且我有以下功能：

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

当我尝试编译这个时，以下错误：

 无法匹配预期类型'a'与推断类型'Double'
`a'是一个刚性类型变量，由
绑定，在Perlin.hs中为`getN'的类型签名：15：12 
在表达式中：5.0 :: Double 
在`getN'的定义中：getN = 5.0 :: Double 
  

据我所知，函数设置为返回在班级中。 Double在这门课上（ http://www.zvon.org/other/ haskell / Outputprelude / Num_c.html ），所以我本以为在这种情况下可以返回一个Double。

有人可以请解释一下这个问题？

解决方案

具有签名 Num a =>预计可用于类 Num 中的任何类型。实现 5.0 :: Double 只适用于一个类型，而不适用于全部类型的类型，因此编译器抱怨。

一个通用函数的例子是：

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


这适用于任何类型是 Num 。它适用于双打，整数和任何其他你想使用的数字。因为它可以有一个通用类型签名，只需要参数在类 Num 中。 （类型为 Num 是必要的，因为函数使用与在其中定义的 * 相乘）。

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

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

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.

Can someone explain this please?

解决方案

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.

An example of a generic function would be:

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

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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