Pretty printing a syntax tree in Haskell

Question

I don't understand this type error:

Couldn't match expected type `[t0]' with actual type `IO ()'
In the return type of a call of `printv'
In a stmt of a 'do' expression: px <- printv x
In the expression:
  do { px <- printv x;
       sep <- print ", ";
       rest <- prints xs;
       return (px : sep : rest) }

From:

data    Value     = IntValue     Int
                  | TruthValue   Bool
                    deriving (Eq, Show)

printv :: Value -> IO()
printv (IntValue   i) = print i
printv (TruthValue b) = print ("boolean" ++ show b)

prints :: [Value] -> [IO()]
prints  []    =  []
prints (x:xs) = do px   <- printv x
                   sep  <- print ", "
                   rest <- prints xs
                   return (px:sep:rest)

It looks to me like every element (px) is converted into an IO() action, and then that is added to a list of the same things, thus producing an [IO()] list.

What am I missing here? Converting it to a list of strings, by removing the print's, works fine.

Answer 1

Take a closer look at the type of your function:

prints :: [Value] -> [IO()]

But if we now take a look at prints [] = [], this can't match, because the type of that one is

prints :: [t] -> [a]

Therefore, you missed using prints [] = return [], to make it work.