haskell1e.md

1E. Input and Output with the IO Monad

import Control.Monad (liftM)
main = do 
  -----------[4]-----------
        --------[3]--------
        ----[1]---- --[2]--
  a <- (liftM read)(getLine) :: IO Int
  print a   -- [5]

1. Recall that read parses a string to an int. read :: String -> Int. We use liftM to 'lift' the read fn to an IO monadic context. Here, (liftM read) :: IO String -> IO Int.
2. We use an action to perform a side-effect of getting the next line from stdin. Here, getLine :: IO String
3. We use vanilla function application to pass an IO String to a function of type: IO String -> IO Int. The result is an IO Int.
4. The do block's arrow <- takes the monadic value IO Int, pulls out Int and "assigns" it to a. Or in terms of lambdas and binds, we open a new lambda (\a -> ... ) with a bound to the integer pulled out of the monadic value on the RHS of the ->.
5. Finally use print a to return a monadic value, print here has type: Int -> IO Int.

IO Function	Description
getLine	Reads a line
readLn	Reads a line and uses read to parse
print	Writes the value of show to the console
putStrLn	Writes a string to the console with a newline

main = (\val1 -> (\val2 -> (\sum -> (print sum))( val1 + val2 )) =<< readLn ) =<< readLn

Example

do 
  line <- getLine   
  let line' = reverse line  
  putStrLn $ "You said " ++ line' ++ " backwards!"  
  putStrLn $ "Yes, you really said" ++ line' ++ " backwards!"

main = do
    a <- liftM (map toUpper) getLine
    putStrLn a

$ runhaskell example.hs
Hello
"HELLO"

---

$ touch foo{1,2,3}.txt

---

Exercises

3. Write a program that accepts two integers and returns the first minus the second.
Hint: you will need to use readLn and print and one or more let statements). Write your program using (a) a do block (b) using lambdas and binds.