Haskell检查空列表

Question

我想有一个函数来检查空列表并告诉列表的内容。
下面的代码，我不能用[]调用tell函数

tell :: (Show a) => [a] -> String  
tell [] = "The list is empty"  
tell (x:[]) = "The list has one element: " ++ show x  
main = do
putStrLn (tell [])

错误是

main.hs:8:12: error:
* Ambiguous type variable `a0' arising from a use of `tell'
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 the first argument of `putStrLn', namely `(tell [])'
In a stmt of a 'do' block: putStrLn (tell [])
In the expression: do putStrLn (tell [])
|
8 |  putStrLn (tell [])
|            ^^^^^^^
exit status 1

如何解决这个问题?

Answer 1

问题是tell []中的[]没有说明该列表中元素的类型。这一点很重要，因为对于非空列表，我们使用show x，而Double和Int的show是不同的。

您可以使用以下命令提供列表的类型:

main = putStrLn (tell ([]:: [Int]))

当然你可以使用不同的类型，如String,[Double],[[Char]]等。