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Home/ Questions/Q 546437
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T10:52:01+00:00 2026-05-13T10:52:01+00:00

Consider the following piece of Haskell code: type Parser a = String -> [(a,

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Consider the following piece of Haskell code:

 type Parser a = String -> [(a, String)]

 item :: Parser Char
 item = \ inp -> case inp of
                 [] -> []
                 (x:xs) -> [(x, xs)]

 ret :: a -> Parser a
 ret v = \ inp -> [(v, inp)]

 parse :: Parser a -> String -> [(a, String)]
 parse p inp = p inp

 pseq :: Parser (Char, Char)
 pseq = do x <- item
           item
           y <- item
           ret (x, y)

 ac = parse pseq "abcdef"

When trying to run the above code in hugs (Version September 2006), I get the following error message:

Type error in explicitly typed binding
*** Term           : pseq
*** Type           : [Char] -> [(([(Char,[Char])],[(Char,[Char])]),[Char])]
*** Does not match : Parser (Char,Char)

And, when I remove my type declaration for “pseq”, I get the following error message:

Unresolved top-level overloading
*** Binding             : pseq
*** Outstanding context : Monad ((->) [Char])
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  1. Editorial Team
    Editorial Team
    2026-05-13T10:52:01+00:00Added an answer on May 13, 2026 at 10:52 am

    Making Parser a Monad is easy, and I think resorting to ListT or StateT is probably overkill here.

    newtype Parser a = Parser (String -> [(a, String)])
        -- make this a distinct type
        -- now the function is "wrapped", though at no run-time cost
    
    instance Monad Parser where
        return = ret    -- you already had it defined
        (>>=) = bind    -- added below
        -- this instance makes Parser a Moand, and lets it work with do notation
    
    item :: Parser Char
    item = Parser $ \ inp -> case inp of
                             [] -> []
                             (x:xs) -> [(x, xs)]
        -- need to "wrap" up the function as a Parser value
    
    ret :: a -> Parser a
    ret v = Parser $ \ inp -> [(v, inp)]
        -- need to "wrap" up the function as a Parser value
    
    bind :: Parser a -> (a -> Parser b) -> Parser b
    bind p f = Parser $ \ s -> case parse p s of
                               [] -> []
                               [(a, s')] -> parse (f a) s'
        -- the only function you were missing
        -- it is worth making sure you understand how it works
    
    parse :: Parser a -> String -> [(a, String)]
    parse (Parser p) inp = p inp
        -- needs to "unwrap" the newtype here
    
    pseq :: Parser (Char, Char)
    pseq = do x <- item
              item
              y <- item
              ret (x, y)
    
    ac = parse pseq "abcdef"
        -- you probably meant pseq here, not seq
    

    Lastly, you use the return type of [(a, String)], so that you can indicate things which can’t parse. But the list only ever has zero or one items in it. You should look into the types Maybe and Either which would likely be clearer in this case.

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