Haskell pre-monadic I/O

Question

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

Answer 1

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