Haskell为什么要“Num x”需要“显示x”？ [英] Haskell why does "Num x" require "Show x"?

问题描述

最近，我使用 LYAH 来看看Haskell。

  foo ::（Num x）=>这是一个快速测试函数。 x  - >字符串
 foo x =显示x ++！ 
  

但是会产生这个错误：

<$ p $ b $ test.hs：2：9：
从上下文中使用`show'
引起的（显示x）（Num x）
由foo :: Num x =>的类型签名绑定。 x - >字符串
在test.hs：1：8-29
可能的修正：
添加（显示x）到
的上下文foo :: Num x =>的类型签名; x - >字符串


但是根据LYAH：

为了加入Num，一个类型必须已经是Show和Eq的朋友。

code> Num 是显示和 Eq 的子集，为什么我需要将类型签名更改为 foo ::（Num x，Show x）=> x - >字符串为此工作？难道不可能推断出 Num 也是可显示的吗？

解决方案

div>

LYAH的信息很旧。 GHC 7.4.1发布说明说：

Num类不再有Eq或Show超类。

您需要编写

  foo ::（Num x，Show x）= > x  - >字符串
  

（事实上， foo 你写不需要 Num x ，所以你可以忽略它以避免不必要的约束。）

Recently I took a look at Haskell, using LYAH.

I was messing around with type classes and wrote this quick test function:

foo :: (Num x) => x -> String
foo x = show x ++ "!"

But that produces this error:

test.hs:2:9:
    Could not deduce (Show x) arising from a use of `show'
    from the context (Num x)
    bound by the type signature for foo :: Num x => x -> String
    at test.hs:1:8-29
    Possible fix:
      add (Show x) to the context of
        the type signature for foo :: Num x => x -> String

But according to LYAH:

To join Num, a type must already be friends with Show and Eq.

So if everything in Num is a subset of Show and Eq, why do I need to change the type signature to foo :: (Num x, Show x) => x -> String for this to work? Shouldn't it be possible to infer that a Num is also Show-able?

解决方案

The information in LYAH is old. The release notes for GHC 7.4.1 say that:

The Num class no longer has Eq or Show superclasses.

You will need to write,

foo :: (Num x, Show x) => x -> String

(In fact, the foo you wrote doesn't require Num x, so you can omit that to avoid an unnecessary constraint.)

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