To make do notation work, you have to
(1) use a data type for Parser (instead of a type synonym) and (2) declare that data type an instance of the Monad typeclass.
Your code should look something like this:
data Parser = Parser (...) instance Monad Parser where return v = ... Parser p >>= f = ...
...
Thanks, I'll try this. I haven't read upto the data declaration and inheritance yet, but I'll try it as soon as I have :-)
(>>=) :: Parser a -> (a -> Parser b) -> Parser b p >>= f = \inp -> case parse p inp of [] -> [] [(v, out)] -> parse (f v) out
What happens if p returns more then one result?
The way Parsers are designed in the book, it can't. Parsers always return a list with one element. The only reason the return value is a list is to use the empty list as an error indication. Arindam