Haskell中seqn函数编译报错求助：代码与错误分析

你写的seqn函数代码：

seqn [] = return []
seqn (act:acts) = do x  <- act
                     xs <- seqn acts
                     return (x::xs)

GHCi给出的错误信息：

seqn.hs:4:30: error:
    • Couldn't match expected type ‘xs’ with actual type ‘a’
      ‘xs’ is a rigid type variable bound by
        an expression type signature:
          forall xs. xs
        at seqn.hs:4:33-34
      ‘a’ is a rigid type variable bound by
        the inferred type of seqn :: Monad m => [m a] -> m [a1]
        at seqn.hs:(1,1)-(4,35)
    • In the first argument of ‘return’, namely ‘(x :: xs)’
      In a stmt of a 'do' block: return (x :: xs)
      In the expression:
        do x <- act
           xs <- seqn acts
           return (x :: xs)
    • Relevant bindings include
        x :: a (bound at seqn.hs:2:22)
        acts :: [m a] (bound at seqn.hs:2:11)
        act :: m a (bound at seqn.hs:2:7)
        seqn :: [m a] -> m [a1] (bound at seqn.hs:1:1)
  |
4 |                      return (x::xs)
  |                              ^
Failed, no modules loaded.

问题原因 #

错误根源在最后一行的x::xs：Haskell里::是类型标注运算符，用来指定表达式的类型；而你实际需要的是列表构造符:（单个冒号），用来把元素x添加到列表xs的头部。你误把单个冒号写成了双冒号，导致GHC将xs解析成一个类型变量，而非之前通过xs <- seqn acts绑定的列表值，从而触发类型不匹配的错误。

修正方案 #

把x::xs改成x:xs即可，修正后的完整代码：

seqn [] = return []
seqn (act:acts) = do 
    x  <- act
    xs <- seqn acts
    return (x:xs)

修正后，函数逻辑符合预期：递归处理动作列表，先执行当前动作得到结果x，再递归处理剩余动作得到结果列表xs，最后将x前置到xs并通过return包装为Monad类型的值，类型签名会被正确推导为seqn :: Monad m => [m a] -> m [a]。

内容的提问来源于stack exchange，提问作者Gergely