[Haskell-beginners] How to run Parser in chapter 8 Graham Hutton Haskell book

Angus Comber anguscomber at gmail.com
Fri Jan 3 15:58:52 UTC 2014

I am reading Chapter 8 of Programming Haskell by Graham Hutton and trying
to run the code in the book.

It seems that there are already defined library implementations of Parser
so I used Parser' and generally ' on the end of each identifier to attempt
to eliminate this problem.

So the code I copied from the book is:

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

return' :: a -> Parser' a
return' v = \x -> [(v, x)]

failure' :: Parser' a
failure' = \inp -> []

-- item doesn't seem to conflict with anything so no '
item :: Parser' Char
item = \inp -> case inp of
                    [] -> []
                    (x:xs) -> [(x,xs)]

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

p :: Parser' (Char, Char)
p = do  x <- item
        y <- item
        return' (x,y)

When run from WinGHCi I get error:

   Couldn't match type `[(Char, String)]' with `Char'
    Expected type: String -> [((Char, Char), String)]
      Actual type: Parser' ([(Char, String)], [(Char, String)])
    In the return type of a call of return'
    In a stmt of a 'do' block: return' (x, y)
    In the expression:
      do { x <- item;
           y <- item;
           return' (x, y) }

458.9 is line at end with return' (x,y)

Also in the chapter was the definition of >>= sequencing.  as in:

(>>=) :: Parser a -> (a -> Parser b) -> Parser b
p >>= f  =  \inp -> case parse p inp of
                      [] -> []
                      [(v,out)] -> parse (f v) out

But I assumed this would already be in standard Haskell library and so I
didn't need to define.  But even when I did add, still get same error.

How do I fix this?
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