Haskell foldr会导致类型错误，而foldl不会

问题描述

我正在通过Haskell从第一原则编程。在折叠列表一章中，练习5f，当我评估时，

  foldr const'a'[1..5] 
  

我得到

然而，

  foldl const'a'[1..5] 
  

我得到'a'。

我得到褶皱是懒惰的， foldr 不会穿过脊椎，而 foldl 却会。但是，即使是在查看foldr和foldl的定义，

  foldr fz [] = z 
 foldr fz（x： xs）= fx（foldr fz xs）
 
 foldl fz [] = z 
 foldl fz（x：xs）= foldl f（fzx）xs 
  

我无法弄清楚为什么会出现这种类型的错误。我猜测编译器是根据<$ c $的类型推断 x （ Num ）的类型因为 const （ Char ），但我看不到它在哪里绘制连接。 c $ c>或 f 并不要求它的两个参数是相同的类型。

有什么想法？

解决方案

确定看看 foldr ::（a - > b - > b ） - > b - > [a] - > b

从右边开始，显然必须有 a 是 Num 和 Enum （因为您使用 [1..5] ）

接下来你传入'a'，所以你有 b〜Char

最后，您的函数有 const - 而const是 const :: a - > b - > a - 请注意您现在必须拥有 a〜b ，因为您统一了

  a  - > b  - > b 
 a  - > b  - > a 
 ^^^^^ 
  

但这当然意味着'a'必须是 Num 其中 Char 的实例值不是......这是你的错误（它是在抱怨1，因为从左边开始，而不是问题变得明显）

---

foldl 在另一侧具有 foldl ::（b - > a - > b） - > b - > [a] - > b

所以现在您再次 a Num ， b 必须再次为 Char ，但现在 const 恰好符合（在类型const中切换 b 和 a ）

I'm working through "Haskell Programming From First Principles". In the chapter on Folding Lists, exercise 5f,

when I evaluate

foldr const 'a' [1..5]

I get

However, with

foldl const 'a' [1..5]

I get 'a'.

I get that the folds are lazy, foldr doesn't traverse the spine and foldl does. But even looking at the definitions of foldr and foldl,

foldr f z []     = z
foldr f z (x:xs) = f x (foldr f z xs)

foldl f z []     = z
foldl f z (x:xs) = foldl f (f z x) xs

I can't figure out why it would have this type error. I'm guessing that the compiler is inferring type of x (Num) based on the type of z (Char), but I can't see where it draws the connection, since const or f doesn't require its two arguments to be the same type.

Any thoughts?

解决方案

ok look at the type of foldr :: (a -> b -> b) -> b -> [a] -> b

Starting from the right you obviously must have a be some instance of Num and Enum (because you use [1..5])

Next you pass in 'a' so you have b ~ Char

Finally you have const for the function - and const is const :: a -> b -> a - note how you now must have a ~ b because you unify

a -> b -> b
a -> b -> a
        ^^^^^

but this of course would mean that 'a' must be a value of an instance of Num which Char is not ... there is your error (it's complaining about the 1 because starting left instead it's the point where the the problem becomes obvious)

---

foldl on the other side has foldl :: (b -> a -> b) -> b -> [a] -> b

so now you have again a some instance of Num, b must again be Char but now const just fits in (just switch b and a in the type of const)

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