Re: [Haskell-cafe] Type signature for function where function parameter is used twice?

FYI f x = g x x can be written: f = join g On 19/07/15 18:51, Clinton Mead wrote:

I guess my question is, how to I type LHS so it is no more specialised as RHS in this case?

On Sun, Jul 19, 2015 at 6:42 PM, Ivan Lazar Miljenovic mailto:ivan.miljenovic@gmail.com> wrote:

My understanding is that when you have "LHS = RHS", you can replace any instance of LHS with RHS but you can't always do it the other way around, as the LHS might be more specialised (as in this case it is).

On 19 July 2015 at 18:08, Clinton Mead mailto:clintonmead@gmail.com> wrote: > Thanks for the CC Ivan, forgot to "reply all". > > I guess I want the passed argument to stay polymorphic long enough to be > instanced inside the call. > > Surely if "f x = g x x" then you should somehow be able to replace "g x x" > with "f x"?! That's basically what I'm getting at. > > > > On Sun, Jul 19, 2015 at 5:54 PM, Ivan Lazar Miljenovic > mailto:ivan.miljenovic@gmail.com> wrote: >> >> (Re-CC'ing Cafe) >> >> I think what's happening here is that when you do "apply f1 f1", the >> two c's and d's don't actually match with each other. >> >> But when you do it with applyBoth, they *have* to match each other. >> >> As a comparison, compare: >> >> :t (read, read) >> (read, read) :: (Read a, Read a1) => (String -> a, String -> a1) >> >> and >> >> :t Control.Monad.join read >> Control.Monad.join (,) read :: Read a => (String -> a, String -> a) >> >> I'm not sure how you could define such a function to do what you're >> requesting; my attempt is this: >> >> applyBoth :: (forall a b a2 b2. arr a b -> arr a2 b2) -> D arr a b -> >> D arr a2 b2 >> applyBoth f = apply f f >> >> but whilst ghci accepts it, I couldn't actually seem to use it (lots >> of errors about being unable to match types). >> >> On 19 July 2015 at 17:15, Clinton Mead mailto:clintonmead@gmail.com> wrote: >> > Thanks for your reply Ivan >> > >> > Here's a full code example ( http://ideone.com/1K3Yhw ): >> >> > >> > --- >> > >> > data D c a b = D (c a b) (c b a) >> > >> > apply :: (c a b -> c a2 b2) -> (c b a -> c b2 a2) -> D c a b -> D c a2 >> > b2 >> > apply f1 f2 (D x y) = D (f1 x) (f2 y) >> > >> > applyBoth f = apply f f >> > >> > f :: Int -> String >> > f = show >> > >> > g :: String -> Int >> > g = read >> > >> > h :: Int -> Int >> > h = (*2) >> > >> > join :: (a -> b) -> (c -> d) -> ((a, c) -> (b, d)) >> > join f g (x, y) = (f x, g y) >> > >> > x1 = D f g >> > >> > f1 = join h >> > >> > y1 = apply f1 f1 x1 -- This compiles >> > y2 = apply g g x1 where g = f1 -- Also works >> > >> > z1 = applyBoth f1 x1 -- This fails >> > >> > main = return () >> > >> > --- >> > >> > You see, with "y1" and "y2", a different "f1" instantiation (is that the >> > right word) of "f1" is chosen. Even when I let "g = f1", each "g" passed >> > to >> > apply is different, the first operates on "Int -> String", the second, >> > "String -> Int". >> > >> > On the face of it, if: >> > >> > f x = g x x >> > >> > Then well, you should be able to replace "g x x" with "f x" wherever you >> > see >> > it. >> > >> > But it seems in this case this doesn't work. Surely Haskell doesn't >> > require >> > me to "repeat myself", and there's a way to write "applyBoth", instead >> > of >> > having to write "apply f f" (note the repetition of f) all the time? >> > >> > I assume there's some type signature I can give "applyBoth f" so it can >> > be >> > used just like "apply f f" but I can't work out what it is. >> > >> > On Sun, Jul 19, 2015 at 10:23 AM, Ivan Lazar Miljenovic >> > mailto:ivan.miljenovic@gmail.com> wrote: >> >> >> >> >> On 19 July 2015 at 10:13, Clinton Mead mailto:clintonmead@gmail.com> wrote: >> >> > Lets say I've got the following data type: >> >> > >> >> > data D c a b = D (c a b) (c b a) >> >> > >> >> > And I define a function to manipulate it: >> >> > >> >> > apply :: (c a b -> c a2 b2) -> (c b a -> c b2 a2) -> D c a b -> D c >> >> > a2 >> >> > b2 >> >> > apply f1 f2 (D x y) = D (f1 x) (f2 y) >> >> > >> >> > This is all fine. But I want a shorter function if (f1 = f2). So I >> >> > write: >> >> > >> >> > applyBoth f = apply f f >> >> > >> >> > I originally thought that if "apply f f" is valid, then logically >> >> > "applyBoth >> >> > f" should also be valid. But it seems that type inference results in >> >> > applyBoth only working for functions "c a a -> c a2 a2". >> >> >> >> applyBoth f = apply f f >> >> >> >> In apply, (f1 :: c a b -> c a2 b2) and (f2 :: c b a -> c b2 a2). >> >> >> >> So for "apply f f" to typecheck, we must have that a ~ b and a2 ~ b2 >> >> (as the above two type signatures must match since f1 = f2). >> >> >> >> So we must have that (f :: c a a -> c a2 a2). >> >> >> >> > >> >> > Is there a way to type "applyBoth" so it works for all functions that >> >> > would >> >> > work simply by repeating them twice in "apply"? >> >> >> >> I'm not sure what you mean; what else would make sense? Can you >> >> provide an example? >> >> >> >> -- >> >> Ivan Lazar Miljenovic >> >> Ivan.Miljenovic@gmail.com mailto:Ivan.Miljenovic@gmail.com >> >> http://IvanMiljenovic.wordpress.com >> > >> > >> >> >> >> -- >> Ivan Lazar Miljenovic >> Ivan.Miljenovic@gmail.com mailto:Ivan.Miljenovic@gmail.com >> http://IvanMiljenovic.wordpress.com > >

-- Ivan Lazar Miljenovic Ivan.Miljenovic@gmail.com mailto:Ivan.Miljenovic@gmail.com http://IvanMiljenovic.wordpress.com

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