Idiomatic Haskell syntax without do-blocks or accumulating parentheses? - haskell

Question

I'm fairly new to Haskell and have been trying to find a way to pass multiple IO-tainted values to a function to deal with a C library. Most people seem to use the <- operator inside a do block, like this:
g x y = x ++ y
interactiveConcat1 = do {x <- getLine;
y <- getLine;
putStrLn (g x y);
return ()}
This makes me feel like I'm doing C, except emacs can't auto-indent. I tried to write this in a more Lispy style:
interactiveConcat2 = getLine >>= (\x ->
getLine >>= (\y ->
putStrLn (g x y) >>
return () ))
That looks like a mess, and has a string of closed parentheses you have to count at the end (except again, emacs can reliably assist with this task in Lisp, but not in Haskell). Yet another way is to say
import Control.Applicative
interactiveConcat3 = return g <*> getLine <*> getLine >>= putStrLn
which looks pretty neat but isn't part of the base language.
Is there any less laborious notation for peeling values out of the IO taint boxes? Perhaps there is a cleaner way using a lift* or fmap? I hope it isn't too subjective to ask what is considered "idiomatic"?
Also, any tips for making emacs cooperate better than (Haskell Ind) mode would be greatly appreciated. Thanks!
John
Edit: I stumbled across https://wiki.haskell.org/Do_notation_considered_harmful and realized that the nested parentheses in the lambda chain I wrote is not necessary. However it seems the community (and ghc implementors) have embraced the Applicative-inspired style using , <*>, etc, which seems to make the code easier to read in spite of the headaches with figuring out operator precedence.

Answer 1

Note: This post is written in literate Haskell. You can save it as Main.lhs and try it in your GHCi.
A short remark first: you can get rid of the semicolons and the braces in do. Also, putStrLn has type IO (), so you don't need return ():
interactiveConcat1 = do
x <- getLine
y <- getLine
putStrLn $ g x y
We're going to work with IO, so importing Control.Applicative or Control.Monad will come in handy:
> module Main where
> import Control.Applicative
> -- Repeat your definition for completeness
> g :: [a] -> [a] -> [a]
> g = (++)
You're looking for something like this:
> interactiveConcat :: IO ()
> interactiveConcat = magic g getLine getLine >>= putStrLn
What type does magic need? It returns a IO String, takes a function that returns an String and takes usual Strings, and takes two IO Strings:
magic :: (String -> String -> String) -> IO String -> IO String -> IO String
We can probably generalize this type to
> magic :: (a -> b -> c) -> IO a -> IO b -> IO c
A quick hoogle search reveals that there are already two functions with almost that type: liftA2 from Control.Applicative and liftM2 from Control.Monad. They're defined for every Applicative and – in case of liftM2 – Monad. Since IO is an instance of both, you can choose either one:
> magic = liftA2
If you use GHC 7.10 or higher, you can also use <$> and <*> without import and write interactiveConcat as
interactiveConcat = g <$> getLine <*> getLine >>= putStrLn
For completeness, lets add a main so that we can easily check this functionality via runhaskell Main.lhs:
> main :: IO ()
> main = interactiveConcat
A simple check shows that it works as intended:
$ echo "Hello\nWorld" | runhaskell Main.lhs
HelloWorld
References
Applicative in the Typeclassopedia
The section "Some useful monadic functions" of LYAH's chapter "For a Few Monads More".

Answer 2

You can use liftA2 (or liftM2 from Control.Monad):
import Control.Applicative (liftA2)
liftA2 g getLine getLine >>= putStrLn