haskell 。 "Could not deduce (t ~ [t]) from the context (Eq t)"

Question

仅仅自学haskell，并不能解决这个问题。这些是代码。

subst :: Eq t=>t->t->[t]->[t]
subst a b [xs]=[if x==a then b else x | x <- xs]

错误如下。

subst.hs:2:46:
    Could not deduce (t ~ [t])
    from the context (Eq t)
      bound by the type signature for
                 subst :: Eq t => t -> t -> [t] -> [t]
      at subst.hs:1:10-29
      ‘t’ is a rigid type variable bound by
          the type signature for subst :: Eq t => t -> t -> [t] -> [t]
          at subst.hs:1:10
    Relevant bindings include
      xs :: t (bound at subst.hs:2:12)
      b :: t (bound at subst.hs:2:9)
      a :: t (bound at subst.hs:2:7)
      subst :: t -> t -> [t] -> [t] (bound at subst.hs:2:1)
    In the expression: xs
    In a stmt of a list comprehension: x <- xs

我猜问题在于 haskell 无法确保 [t] 中的元素与 t 匹配。我不知道。我想知道如何解决这个问题。

Answer 1

你只有一个[..]包装到许多:

subst :: Eq t=>t->t->[t]->[t]
subst a b xs = [if x==a then b else x | x <- xs]

这只是xs不是[xs]

原因很简单:如果你写 [xs]你告诉 Haskell 期待一个包含单个条目的列表 xs它会尝试模式匹配它 - 之后你告诉它提取值 x来自xs ( x <- xs ) 告诉 Haskell xs本身必须是某种列表。所以最后它会假设 t成为一些 list t ~ [s] 。但然后你继续检查 x == a在这里x :: s和a :: t以现在的错误结束 [s] ~ t ~ s .