On Dec 19, 2006, at 10:11 PM, Dan Weston wrote:
instance CommandFunction (Sh st ()) st where ^ I think your first argument (on which the second has a functional dependence) does not determine the second argument, since it makes use of st in the first argument. This strikes me as a likely place to begin.
No, I'm pretty sure this isn't a problem. The second argument is determined _because_ it is mentioned in the first. The functional dependencies and instance declarations work, as long as I can make the compiler accept them. They are only being rejected by the termination-checking part of the algorithm. That said, I'm open to the idea of reformulating these instances. In fact, I don't really like the fact that I need FDs. It seems to me that I should somehow be able to eliminate the second argument altogether and thus the FD, but I can't seem to figure it out.
Dan
Robert wrote:
Fellow Haskellers, I have a package that uses some light typeclass hackery to automaticly build parsing algorithms based on the type of a function. I was recently informed that my package doesn't compile on GHC 6.6 due to the new restrictions on FD resolution; in particular I have instance declarations which fail the coverage condition. I can use undecidable instances to make the package compile again, but I'd prefer not to if I can avoid it. class CommandFunction f st | f -> st where parseCommand :: String -> f -> CommandParser st commandSyntax :: f -> [Doc] instance CommandFunction (Sh st ()) st where
parseCommand wbc m str = -- list monad do (x,[]) <- runRegex (maybeSpaceBefore (Epsilon (CompleteParse m))) str return x commandSyntax _ = [] instance CommandFunction r st => CommandFunction (Int -> r) st where parseCommand = doParseCommand Nothing intRegex id commandSyntax f = text (show intRegex) : commandSyntax (f undefined) instance CommandFunction r st => CommandFunction (Integer -> r) st where parseCommand = doParseCommand Nothing intRegex id commandSyntax f = text (show intRegex) : commandSyntax (f undefined)
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