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0 pointsZak12y ago0 comments

A thing that people are often unaware of when trying to understand how the IO type works in Haskell is that you cannot define the IO type in Haskell.

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4 comments · 2 top-level

kenko12y ago· 2 in thread

Sure you can: http://hackage.haskell.org/packages/archive/ghc-prim/0.2.0.0...

You can't define RealWorld, though, nor (IIRC) is IO really a state monad.

ZakOP12y ago

I used "define" imprecisely.

You cannot, so far as I know, implement `>>=` for the IO type defined in the above link within Haskell, which means you can't actually use it to do IO.

kenko12y ago

http://www.haskell.org/ghc/docs/latest/html/libraries/base/s...

applicative12y ago

Here's another definition -- by continuation -- from Hug's Prelude (http://cpansearch.perl.org/src/AUTRIJUS/Language-Haskell-0.0...) :

    newtype IO a = IO ((a -> IOResult) -> IOResult)

    data IOResult 
      = Hugs_ExitWith    Int
      | Hugs_Catch       IOResult (Exception -> IOResult) (Obj -> IOResult)
      | Hugs_ForkThread  IOResult IOResult
      | Hugs_DeadThread
      | Hugs_YieldThread IOResult
      | Hugs_Return      Obj
      | Hugs_BlockThread (Obj -> IOResult) ((Obj -> IOResult) -> IOResult)

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