On Thursday, 10. June 2010 22:10:08 Maciej Piechotka wrote: Wow! this is somewhat above my level. I guess I need to go back to the books. I'll document my ignorance nontheless.
data Named a = Named String a
instance Functor Named where f `fmap` (Named s v) = Named s (f v)
okay so far
instance Applicative Named where pure x = Named "" x (Named s f) <*> (Named t v) = Named (s ++ "(" ++ t ++ ")") (f v)
Applicative. Need to study that Control.Applicative (<*>) :: Applicative f => f (a -> b) -> f a -> f b So in our case the Applicative is a "Named". When I apply a Named to a function, then I get a function between the corresponding Named types. When I pass it an Int->Char function, I get a Named Int -> Named Char function. But here it is applied to another Named ... is that the (a->b)? Puzzeled.
instance Eq a => Eq (Named a) where (Named _ x) == (Named _ y) = x == y
instance Show (Named a) where show (Named s _) = s
Understood.
namedPure :: Show a => a -> Named a namedPure x = Named (show x) x
When I can show something I can always name it so its name is what 'show' would return. Okay I guess I got it. This turns a "showable" into a Named.
test :: Num a => (a -> a) -> (a -> a) -> (a -> a) -> [String] test f g h = do [f', g', h'] <- permutations [Named "f" f, Named "g" g, Named "h" h]
According to Hoogle permutations should be in Data.List. Mine (GHCI 6.8.2) does not seem to have it. Seems to have something to do with "base", whatever that is.
guard $ namedPure 42 == f' <*> g' <*> h' <*> namedPure 42
Ah, the 42 needs namedPure. Again this <*> operator... I believe the whole thing is using a List Monad.
return $ show f' ++ " . " ++ show g' ++ " . " ++ show h'
I wonder if the thing returns just one string or a list of strings. I guess "return" cannot return anything more unwrapped than a List, so it must be a List. But does it contain just the first match or all of them? All of them! And how many brackets are around them? -- Martin