Question

我只是想在Haskell中编写我能想象到的最简单的函数，并得到了这个错误信息。奇怪的是，当我尝试评估myHead的空列表时，它才出现。我做错了什么？

module Main 
  where

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

main = do
print (myHead [])

当我从文件中运行它时，我得到了这个输出：

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

$array = explode(",","3,4,5"); if (in_array(3, $array)) { echo 'in array'; } else { echo 'not in array';没有任何问题，如果使用的话，你会遇到同样的问题：

myHead

此处的问题是main = do print Nothing和Nothing对于任何myHead []都具有多态类型Maybe a。然后，调用a来写入该值。为此，print必须要求print可转换为字符串：它通过要求Maybe a执行此操作，而Show (Maybe a)则需要Show a。

但是，没有Show a的通用实例：编译器现在需要知道a之前的内容才能将其转换为字符串。

请注意

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

解决方案是为代码使用具体类型

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

没有输入空列表：Haskell

1 个答案: