haskell - 无法从文字 ‘0’ 中推断出 (Eq i)

Question

我是 Haskell 的新手，正在尝试编写一个类似于 take 的函数，即从指定列表中返回指定数量的项目。我的代码如下:

take' :: (Num i, Ord a) => i -> [a] -> [a]
take' 0 _ = []
take' _ [] = []
take' n (x:xs) = x : take' (n - 1) xs

但是，当我尝试编译它时，我收到以下错误:

Could not deduce (Eq i) arising from the literal ‘0’
from the context (Num i, Ord a)
  bound by the type signature for
             take' :: (Num i, Ord a) => i -> [a] -> [a]
  at recursion.hs:1:10-42
Possible fix:
  add (Eq i) to the context of
    the type signature for take' :: (Num i, Ord a) => i -> [a] -> [a]
In the pattern: 0
In an equation for ‘take'’: take' 0 _ = []

我认为该错误是由于 Haskell 无法将 0 识别为类类型 Num 的成员而引起的，但我不确定。任何人都可以向我解释错误并告诉我如何解决它。

Answer 1

针对文字数字的模式匹配脱糖到相等性检查。所以

take' 0 _ = []

成为

take' x _ | x == 0 = []

(其中 x 被选为子句中其他任何地方未提及的变量)。所以要支持这个模式，你需要拿的东西的数量不仅仅是一个Num，还要支持(==)!这就是错误的这一部分所说的:

Could not deduce (Eq i) arising from the literal ‘0’
In the pattern: 0
In an equation for ‘take'’: take' 0 _ = []

您可以只采用 GHC 在错误中建议的修复:

Possible fix:
  add (Eq i) to the context of
    the type signature for take' :: (Num i, Ord a) => i -> [a] -> [a]

因此:

take' :: (Eq i, Num i, Ord a) => i -> [a] -> [a]

之后您可以考虑是否需要 Ord a 约束。 =)