Question

我使用x＆lt; - getLine获取值，如何检查x是否可以解释为整数？

Answer 1

do x <- getLine
   case filter (\(_,s) -> s == "") (reads x :: [(Int, String)]) of
      [] -> putStrLn "x cannot be parsed as an Int"
      (xAsInt, _) : _
        -> putStrLn (concat ["x can be parsed as an Int, *and* its Int value is ",
                             show xAsInt])

Answer 2

查看Data.Char.isNumber。

Answer 3

Haskell: Check if integer, or check type of variable

Answer 4

您可以创建一个maybeIO函数，在IO中执行catch操作，如果成功则返回Just操作的结果，或Nothing如果发生异常然后，您可以使用readLn代替getLine + reads，maybeIO将任何例外转换为Nothing。

import Control.Monad (liftM)

maybeIO :: IO a -> IO (Maybe a)
maybeIO f = catch (liftM Just f) (const $ return Nothing)

main = do
  i <- maybeIO (readLn :: IO Int)
  print i

getLine并且是值整数

4 个答案: