The below code works fine but I have the feeling that there must be something more idiomatic way to do this within the do-notation without having to resort to the two liftMs?

main = do
  n <- (liftM read (liftM head getArgs)) :: IO Int
  putStrLN . show $ sum [1..n]

If I do:

main = do
  args <- getArgs
  arg1 <- head args
  n <- read arg1 :: IO Int
  putStrLn . show $ sum [1..n]

I get:

Expected type: String
  Actual type: Char
In the first Argument of 'read', namely 'arg1'

which I do not understand. <- getArgs returns IO [String], so <- head args should return IO String, but it seems to return IO Char?! (as per error message).

So my two questions are:

  1. what is the most idiomatic way do do this
  2. what's wrong in the second version?
  • 1
    \$\begingroup\$ main = do [n] <- getArgs; print $ sum [1..read n] \$\endgroup\$
    – Gurkenglas
    Nov 24, 2017 at 21:29

2 Answers 2


I have the Feeling that there must be something more idiomatic way to do this within the do-notation withouth having to resort to the two liftMs?

Yes, a single one. liftM is fmap nowadays. It stems from a time when Functor wasn't a superclass for Monad. The Functor laws state that

fmap f . fmap g = fmap (f . g)

Therefore, we can simplify

liftM read (liftM head getArgs)


fmap (read . head) getArgs

We end up with

main = do
  n <- fmap (read . head) getArgs :: IO Int
  putStrLN . show $ sum [1..n]

Also, putStrLn . show is print. If we introduce a small helper function, the code gets more readable:

readFirstElement :: Read a => [String] -> a 
readFirstElement = read . head

sumNaturalUpTo :: Int -> Int
sumNaturalUpTo n = sum [1..n]

main :: IO ()
main = do
  n <- readFirstElement <$> getArgs
  print $ sumNaturalUpTo n

Note that sum [1..n] is n * (n + 1) `div` 2, though, so we can optimize sumNaturalUpTo to

sumNaturalUpTo n = n * (n + 1) `div` 2

Alternatively, we can write

main :: IO ()
main = do
   args <- getArgs
   print $ sumNaturalUpTo $ readFirstElement args

Which might be the easiest one to read.

what's wrong in the second version?

args is not a IO … value.

  • \$\begingroup\$ So (<$>) = fmap? \$\endgroup\$
    – A Sz
    Nov 24, 2017 at 20:17
  • \$\begingroup\$ @ASz Right. I'm sure I had that in a draft., sorry. \$\endgroup\$
    – Zeta
    Nov 25, 2017 at 8:28

This is the code you're looking for:

main = do
  args <- getArgs
  let arg1 = head args
  let n = (read arg1) :: Int
  putStrLn . show $ sum [1..n]

Think of the <- operator as "extract". It "pulls" a value out of a monad.

But once you've pulled args out of the IO monad, you've now got a pure list of strings. It's not in the IO monad, so you can't extract from it. let instead just does a pure binding like you're used to in non-monadic code, which is what you need here.


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