Question

我有以下简单的代码来查找转换为int的字符串是否有大于127的字符：

isbinary ss = do
    if any (>127) ss 
    then return "True"
    else return "False"

mystr = "this is a test"
main = do 
    print (isbinary mystr)

但是，我遇到以下错误：

$ runghc isbinary.hs

isbinary.hs:9:9: error:
    • Ambiguous type variable ‘m0’ arising from a use of ‘print’
      prevents the constraint ‘(Show (m0 [Char]))’ from being solved.
      Probable fix: use a type annotation to specify what ‘m0’ should be.
      These potential instances exist:
        instance (Show a, Show b) => Show (Either a b)
          -- Defined in ‘Data.Either’
        instance Show a => Show (Maybe a) -- Defined in ‘GHC.Show’
        instance (Show a, Show b) => Show (a, b) -- Defined in ‘GHC.Show’
        ...plus 14 others
        ...plus 24 instances involving out-of-scope types
        (use -fprint-potential-instances to see them all)
    • In a stmt of a 'do' block: print (isbinary mystr)
      In the expression: do { print (isbinary mystr) }
      In an equation for ‘main’: main = do { print (isbinary mystr) }

isbinary.hs:9:16: error:
    • No instance for (Num Char) arising from a use of ‘isbinary’
    • In the first argument of ‘print’, namely ‘(isbinary mystr)’
      In a stmt of a 'do' block: print (isbinary mystr)
      In the expression: do { print (isbinary mystr) }

print $ isbinary mystr也不起作用。

问题出在哪里，如何解决？谢谢。

Answer 1

问题在于Char不是数字类型（不同于C语言），这就是No instance for (Num Char) arising from a use of ‘isbinary'试图告诉您的。您需要对其进行显式转换。让我们搜索-我在Hoogle中输入了Char -> Int。第二个结果ord是我们想要的。

import Data.Char (ord)

isbinary :: String -> Bool
isbinary s = any (\c -> ord c > 127) s

要使意图更清晰，您可以使用isAscii函数

import Data.Char (isAscii)

isbinary :: String -> Bool
isbinary s = any (\c -> not (isAscii c)) s

稍微清理一下代码（我们通常是eta-reduce并使用camelCase名称）

import Data.Char (isAscii)

isBinary :: String -> Bool
isBinary = any (not . isAscii)

另一种可能的解决方案是编写isBinary = any (> '\127')。

查找是否有int的char> 127的函数

1 个答案: