How to print the last element of a list in Haskell?

Question

I'm doing some Haskell exercises from here: https://wiki.haskell.org/99_questions/Solutions/1 where it shows the solution but it does not show how to print the result

I tried

myLast :: [a] -> IO ()
myLast [] = error "No end for empty lists!"
myLast [x] = print x
myLast (_:xs) = myLast xs


main :: IO ()
main = myLast [1,2,3]

I got

/workspaces/hask_exercises/exercises/src/Lib.hs:10:14: error:
    * No instance for (Show a) arising from a use of `print'
      Possible fix:
        add (Show a) to the context of
          the type signature for:
            myLast :: forall a. [a] -> IO ()
    * In the expression: print x
      In an equation for `myLast': myLast [x] = print x

how can I print the last element?

I think somehow the type of the elements in the list must implement Show, but I don't know exactly how to force that.

Answer 1

Better have

myLast :: [a] -> a
myLast [] = error "No end for empty lists!"
myLast [x] = x -- print x
myLast (_:xs) = myLast xs


main :: IO ()
main = print $ myLast [1,2,3]

This keeps our "business logic" function myLast pure, which is then used in the I/O segment of our program main using a built-in IO primitive, in this case print.

Answer 2

The way to require that the list elements belong to Show is to add that constraint to the type signature:

myLast :: Show a => [a] -> IO ()