非常基本的任务使用foldr [英] Very basic task using foldr

问题描述

  foldr（:) [] 1我只是在一些讲义中使用了一个非常简单的示例2 
  

期待结果

  [1,2] 
  

但是，我收到一个错误。每当我尝试使用++或：作为给予foldr的函数时，我都会收到一个错误。

显然我犯了一些非常明显的错误，但我似乎仍然无法找到它。

任何人都可以帮忙吗？

解决方案

code> foldr 通过传递两个参数 1 和 2 就像一个可变参数函数>而不是 [1,2] 。

遇到这种麻烦时，只需检查函数的类型。您可以在GHCi中做到这一点：

  Prelude> ：t foldr 
 foldr ::（a  - > b  - > b） - > b  - > [a]  - > b 
  

所以你看到第一个参数应该是一个函数（a - > b - > b）。你使用了（：），这就是o.k.您也可以检查部分应用函数的类型：

  Prelude> ：t（:) 
（:) :: a  - > [a]  - > [a] 
  

用<$ $替换 b c $ c> [a] 给了我们：

  Prelude> ：t foldr（:) 
 foldr（:) :: [a]  - > [a]  - > [a] 
  

接下来，您给了 [] 作为基本案例。

  Prelude> ：t foldr（:) [] 
 foldr（:) [] :: [a]  - > [a] 
  

所以结果函数的类型是 [a] - > ; [α] 。你应该怎么做？你必须通过一个列表来获取列表！传递参数列表 [1，2] ：

  Prelude> ：t foldr（:) [] [1,2] 
 foldr（:) [] [1,2] :: Num a => [b] 
  

被类型检查器接受并产生结果：

  Prelude> foldr（:) [] [1,2] 
 [1,2] 
  

I希望这有助于类型 - 调试你的程序...

I just used a very simple example used in some lecture notes with my ghci:

 foldr (:) [] 1 2

expecting the result

   [1,2]

However, I get an error. I get an error everytime when I try to use ++ or : as the function given to foldr.

Apparently I am making some pretty obvious mistake but I still cannot seem to find it.

Can anyone help?

解决方案

You used foldr like a variadic function by passing it two arguments 1 and 2 instead of [1, 2].

When you run into trouble like that, just check the function's type. You can do that in GHCi:

Prelude> :t foldr
foldr :: (a -> b -> b) -> b -> [a] -> b

So you see that the first argument should be a function (a -> b -> b). You used (:) for that, which is o.k. You can check the type of the partially applied function as well:

Prelude> :t (:)
(:) :: a -> [a] -> [a]

Substituting b with [a] gives us:

Prelude> :t foldr (:)
foldr (:) :: [a] -> [a] -> [a]

Next, you gave [] as a base case.

Prelude> :t foldr (:) []
foldr (:) [] :: [a] -> [a]

So the resulting function is of type [a] -> [a]. What should you make of this? You have to pass it a list to get a list back! Passing the argument list [1, 2]:

Prelude> :t foldr (:) [] [1,2]
foldr (:) [] [1,2] :: Num a => [a]

Is accepted by the type checker and yields the result:

Prelude> foldr (:) [] [1,2]
[1,2]

I hope this helps type-debugging your programs...

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