为什么不能改变使用任何Integral的类型？ [英] Why can't the type of take be changed to use any Integral?

问题描述

  take :: Int  - > [a]  - > [a] 
 genericTake :: Integral i =>我 - > [a]  - > [a] 
  

我读过这个不方便的类型取是由于历史原因，而改变它可能会导致一些代码被破坏。

但我不能替换 take 与 genericTake 无处不破？问题是什么？

解决方案

打破案例

  genericTake :: Integral i =>我 - > [a]  - > [a] 
 genericTake n xs = take（fromIntegral n）xs 
 
 class Foo a where 
 bar :: a  - >字符串
 
实例Foo Int其中
 bar _ =int
 
 foo :: String  - > [a]  - > [a] 
 foo ns xs = let y = read ns 
z = bar y 
 in take y xs 
  

这将为 genericTake 打破。

 没有使用'bar'引起的（Foo i0）实例
类型变量'i0'不明确
  

这是一个熟悉的例子，但您可以理解某种类型的推论，它出现在第一个参数中，假设它是 Int ，现在当您将类型更改为 Integral i =>我上面可能会出现一些问题。

take :: Int -> [a] -> [a]
genericTake :: Integral i => i -> [a] -> [a]

I have read that the inconvenient type of take is due to historical reasons, and that changing it could cause some code to break.

But can't I replace take with genericTake everywhere without breaking anything? What's the problem?

解决方案

A breaking case

genericTake :: Integral i => i -> [a] -> [a]
genericTake n xs = take (fromIntegral n) xs

class Foo a where
   bar :: a -> String

instance Foo Int where
   bar _ = "int" 

foo :: String -> [a] -> [a]
foo ns xs = let y = read ns
                z = bar y
            in take y xs

This will break for genericTake.

No instance for (Foo i0) arising from a use of `bar'
    The type variable `i0' is ambiguous

This is a cooked up example but you can understand some type inference occuring on the first argument of take where it is assumed that it is Int, now when you change type to Integral i => i some problems might occur as above.

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