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

You probably want:

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

Assuming that you're following Graham Hutton's book.

Note that this definition won't actually compile; you probably need a Monad instance and a newtype to get this to work properly (see http://www.cs.nott.ac.uk/~gmh/Parsing.lhs for a working version of the same code). PRS: That explains it then. [(v, out)] -> parse (f v) out caused a type mismatch. Cheers, Paul