Answer by Daniel Fischer for Confused about custom data types in Haskell

You need a Show instance to convert the type to a printable representation (a String). The easiest way to obtain one is to add

deriving Show

to the type definition.

data Person = Person {     first_name :: String,     last_name :: String,       age :: Int }      deriving (Eq, Ord, Show)

to get the most often needed instances.

If you want a different Ord instance, as suggested in the comments, instead of deriving that (keep deriving Eq and Show unless you want different behaviour for those), provide an instance like

instance Ord Person where    compare p1 p2 = case compare (age p1) (age p2) of                      EQ -> case compare (last_name p1) (last_name p2) of                              EQ -> compare (first_name p1) (first_name p2)                              other -> other                      unequal -> unequal

or use pattern matching in the definition of compare if you prefer,

    compare (Person first1 last1 age1) (Person first2 last2 age2) =        case compare age1 age2 of          EQ -> case compare last1 last2 of                  EQ -> compare first1 first2                  other -> other          unequal -> unequal

That compares according to age first, then last name, and finally, if needed, first name.