Haskell MonadIO与IO

Question

两者之间有什么区别？

recompile :: MonadIO m => Bool -> m Bool
recompile :: Bool -> IO Bool

Answer 1

类型forall m. MonadIO m => Bool -> m Bool和Bool -> IO Bool是同构的：

{-# LANGUAGE RankNTypes #-}
import Control.Monad.IO.Class

from :: (forall m. MonadIO m => Bool -> m Bool) -> (Bool -> IO Bool)
from act = act

to :: (Bool -> IO Bool) -> (forall m. MonadIO m => Bool -> m Bool)
to act = liftIO . act

Not all types involving IO are isomorphic to a version that replaces IO with a constrained m, though.

通常，当人们觉得通过调用代码分散MonadIO m会很烦人，或者需要使用其他m ~ IO样式约束时，通常使用liftIO而不是mtl m;并在简化API时使用m ~ IO而不是MonadIO m（其中MonadIO m会增加复杂性，因此不受欢迎），或者在处理派生或异常处理时（其中{{1} }由于链接问题中讨论的原因而无法实现。