[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
item
y <- item
return' (x,y)
When run from WinGHCi I get error:
prog_haskell.hs:458:9:
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;
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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