Re: [Haskell-cafe] type class constraints headache

Thanks, it did! (For the record, here is a paraphrase of what first confused me -- undefined was not the problem).

enumerateMethodNames :: [String] enumerateMethodNames = map fst methodsNoConstr --enumerateMethodNames = map fst methodsConstr

methodsConstr :: (Ord a) => [(String, [a] -> Int)] methodsConstr = [ ("method", methodConstr )] where methodConstr :: (Ord a) => [a] -> Int methodConstr xs = length . sort $ xs

methodsNoConstr :: [(String, [a] -> Int)] methodsNoConstr = [ ("method", methodNoConstr )] where methodNoConstr :: [a] -> Int methodNoConstr = length

--First enumerateMethodNames works as expected, second does not compile.

2010/3/4 Ryan Ingram

Perhaps this thought exercise will make things clear:

...

class Show a => Foo a where toFoo :: String -> a

...

foos :: (Foo a) => [(String, a)] foos = map (\f -> (show f, f)) [toFoo "a", toFoo "b", toFoo "c"]

...

data Foo1 = Foo1 instance Show Foo1 where show _ = "1" instance Foo Foo1 where toFoo _ = Foo1 data Foo2 = Foo2 instance Show Foo2 where show _ = "2" instance Foo Foo2 where toFoo _ = Foo2

...

exercise :: [String] exercise = map fst foos

Exercise for the reader: what should the contents of "exercise" be?

Keep in mind that your question is exactly the same as this one, from the compiler's point of view.

-- ryan

On Wed, Mar 3, 2010 at 10:48 PM, Marcus Uneson wrote:

...

Thanks. I realize there are many ways to make it compile. However, I am trying to understand the mechanism behind -- why does the first example compile and what constraints does enumerateMethodNames add on a (which it does not inspect)?