未输入空列表：Haskell

Question

我只是尝试在 Haskell 中编写我能想象到的最简单的可能函数，并收到此错误消息。神奇的是，它只在我尝试评估 myHead 以获得空列表时出现。我做错了什么？

module Main 
  where

myHead :: [a] -> Maybe a
myHead [] = Nothing
myHead (x:_) = Just x

main = do
print (myHead [])

当我从一个文件运行它时，我得到这个输出：

main.hs:15:1: error:
  • Ambiguous type variable ‘a0’ arising from a use of ‘print’
    prevents the constraint ‘(Show a0)’ from being solved.
    Probable fix: use a type annotation to specify what ‘a0’ should be.
 These potential instances exist:
   instance Show Ordering -- Defined in ‘GHC.Show’
   instance Show Integer -- Defined in ‘GHC.Show’
   instance Show a => Show (Maybe a) -- Defined in ‘GHC.Show’
    ... plus 22 others
     ...plus 12 instances involving out-of-scope types
    (use -fprint-potential-instances to see them all)
 • In a stmt of a 'do' block: print (myHead [])
  In the expression: do { print (myHead []) }
  In an equation for ‘main’: main = do { print (myHead []) }
  <interactive>:3:1: error:
   • Variable not in scope: main
   • Perhaps you meant ‘min’ (imported from Prelude)

Answer 1

myHead 没有问题，如果你用了也会遇到同样的问题：

main = do
   print Nothing

这里的问题是Nothing 和myHead [] 对于任何a 都有一个多态类型Maybe a。然后，调用print 来写入该值。为此，print 必须要求 Maybe a 可转换为字符串：它通过要求 Show (Maybe a) 来做到这一点，而 Show a 又要求 Show a。

但是，Show a 没有通用实例：编译器现在需要知道 a 是什么，然后才能将其转换为字符串。

注意这一点

print (Just 3 :: Maybe Int) -- OK
print (Just id :: Maybe (Int->Int)) -- Not OK! Functions can not be printed

解决方案是为您的代码使用具体类型

main = do
   print (myHead [] :: Maybe Int) -- or any other showable type