Question

{-# LANGUAGE DeriveDataTypeable #-}

import Control.Exception
import Data.Typeable

data MyException = MyException String deriving (Show, Typeable)

instance Exception MyException

myExToString :: MyException -> String
myExToString (MyException msg) = msg

t :: ()
t = throw $ MyException "Msg"

main = catch (return t) (\e -> putStrLn $ myExToString e)

为什么我的程序不打印"Msg"？

更新

我更改了代码：

io :: IO ()
io = catch (return t) (\e -> putStrLn $ myExToString e)

main = io >>= print

但我的代码仍然没有抓住MyException？为什么呢？

Answer 1

因为Haskell是惰性的，并且你从不使用t的结果，所以它永远不会被评估，因此不会抛出异常。

Answer 2

关于“以下代码为何不打印例外？”：

io :: IO ()
io = catch (return t) (\e -> putStrLn $ myExToString e)

main = io >>= print

此处，print导致在强制t进行评估时抛出异常，但当时没有catch。请尝试改为：

io :: IO ()
io = catch (return t >>= print) (\e -> putStrLn $ myExToString e)
    -- or simply:  catch (print t) (\e -> ....)
main = io

为什么我的代码没有捕获异常？

2 个答案: