Haskell pre-monadic I/O

HaskellIoMonads

Haskell Problem Overview


I wonder how I/O were done in Haskell in the days when IO monad was still not invented. Anyone knows an example.

Edit: Can I/O be done without the IO Monad in modern Haskell? I'd prefer an example that works with modern GHC.

Haskell Solutions


Solution 1 - Haskell

Before the IO monad was introduced, main was a function of type [Response] -> [Request]. A Request would represent an I/O action like writing to a channel or a file, or reading input, or reading environment variables etc.. A Response would be the result of such an action. For example if you performed a ReadChan or ReadFile request, the corresponding Response would be Str str where str would be a String containing the read input. When performing an AppendChan, AppendFile or WriteFile request, the response would simply be Success. (Assuming, in all cases, that the given action was actually successful, of course).

So a Haskell program would work by building up a list of Request values and reading the corresponding responses from the list given to main. For example a program to read a number from the user might look like this (leaving out any error handling for simplicity's sake):

main :: [Response] -> [Request]
main responses =
  [
    AppendChan "stdout" "Please enter a Number\n",
    ReadChan "stdin",
    AppendChan "stdout" . show $ enteredNumber * 2
  ]
  where (Str input) = responses !! 1
        firstLine = head . lines $ input
        enteredNumber = read firstLine 

As Stephen Tetley already pointed out in a comment, a detailed specification of this model is given in chapter 7 of the 1.2 Haskell Report.


> Can I/O be done without the IO Monad in modern Haskell?

No. Haskell no longer supports the Response/Request way of doing IO directly and the type of main is now IO (), so you can't write a Haskell program that doesn't involve IO and even if you could, you'd still have no alternative way of doing any I/O.

What you can do, however, is to write a function that takes an old-style main function and turns it into an IO action. You could then write everything using the old style and then only use IO in main where you'd simply invoke the conversion function on your real main function. Doing so would almost certainly be more cumbersome than using the IO monad (and would confuse the hell out of any modern Haskeller reading your code), so I definitely would not recommend it. However it is possible. Such a conversion function could look like this:

import System.IO.Unsafe

-- Since the Request and Response types no longer exist, we have to redefine
-- them here ourselves. To support more I/O operations, we'd need to expand
-- these types

data Request =
    ReadChan String
  | AppendChan String String
    
data Response =
    Success
  | Str String
  deriving Show

-- Execute a request using the IO monad and return the corresponding Response.
executeRequest :: Request -> IO Response
executeRequest (AppendChan "stdout" message) = do
  putStr message
  return Success
executeRequest (AppendChan chan _) =
  error ("Output channel " ++ chan ++ " not supported")
executeRequest (ReadChan "stdin") = do
  input <- getContents
  return $ Str input
executeRequest (ReadChan chan) =
  error ("Input channel " ++ chan ++ " not supported")

-- Take an old style main function and turn it into an IO action
executeOldStyleMain :: ([Response] -> [Request]) -> IO ()
executeOldStyleMain oldStyleMain = do
  -- I'm really sorry for this.
  -- I don't think it is possible to write this function without unsafePerformIO
  let responses = map (unsafePerformIO . executeRequest) . oldStyleMain $ responses
  -- Make sure that all responses are evaluated (so that the I/O actually takes
  -- place) and then return ()
  foldr seq (return ()) responses

You could then use this function like this:

-- In an old-style Haskell application to double a number, this would be the
-- main function
doubleUserInput :: [Response] -> [Request]
doubleUserInput responses =
  [    AppendChan "stdout" "Please enter a Number\n",    ReadChan "stdin",    AppendChan "stdout" . show $ enteredNumber * 2  ]
  where (Str input) = responses !! 1
        firstLine = head . lines $ input
        enteredNumber = read firstLine 
  
main :: IO ()
main = executeOldStyleMain doubleUserInput

Solution 2 - Haskell

> I'd prefer an example that works with modern GHC.

For GHC 8.6.5:

import Control.Concurrent.Chan(newChan, getChanContents, writeChan) 
import Control.Monad((<=<))

type Dialogue = [Response] -> [Request]
data Request  = Getq | Putq Char
data Response = Getp Char | Putp

runDialogue :: Dialogue -> IO ()
runDialogue d =
  do ch <- newChan
     l <- getChanContents ch
     mapM_ (writeChan ch <=< respond) (d l)

respond :: Request -> IO Response
respond Getq     = fmap Getp getChar
respond (Putq c) = putChar c >> return Putp

where the type declarations are from page 14 of How to Declare an Imperative by Philip Wadler. Test programs are left as an exercise for curious readers :-)

If anyone is wondering:

 -- from ghc-8.6.5/libraries/base/Control/Concurrent/Chan.hs, lines 132-139
getChanContents :: Chan a -> IO [a]
getChanContents ch
  = unsafeInterleaveIO (do
        x  <- readChan ch
        xs <- getChanContents ch
        return (x:xs)
    )

yes - unsafeInterleaveIO does make an appearance.

Solution 3 - Haskell

@sepp2k already clarified how this works, but i wanted to add a few words

> I'm really sorry for this. I don't think it is possible to write this function without unsafePerformIO

Of course you can, you should almost never use unsafePerformIO http://chrisdone.com/posts/haskellers

I'm using slightly different Request type constructor, so that it does not take channel version (stdin / stdout like in @sepp2k's code). Here is my solution for this:

(Note: getFirstReq doesn't work on empty list, you would have to add a case for that, bu it should be trivial)

data Request = Readline
             | PutStrLn String

data Response = Success
              | Str String

type Dialog = [Response] -> [Request]


execRequest :: Request -> IO Response
execRequest Readline = getLine >>= \s -> return (Str s)
execRequest (PutStrLn s) = putStrLn s >> return Success


dialogToIOMonad :: Dialog -> IO ()
dialogToIOMonad dialog =
    let getFirstReq :: Dialog -> Request
        getFirstReq dialog = let (req:_) = dialog [] in req

        getTailReqs :: Dialog -> Response -> Dialog
        getTailReqs dialog resp =
            \resps -> let (_:reqs) = dialog (resp:resps) in reqs
    in do
        let req = getFirstReq dialog
        resp <- execRequest req
        dialogToIOMonad (getTailReqs dialog resp)

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Solution 3 - HaskellTomatosoupView Answer on Stackoverflow