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

[Clinton Mead]

Avoiding the name class for the moment by qualifying, it seems:

applyBoth = Control.Monad.join apply

has the same issue

On Sun, Jul 19, 2015 at 7:03 PM, Ivan Lazar Miljenovic <
ivan.miljenovic at gmail.com> wrote:

> On 19 July 2015 at 18:51, Clinton Mead <clintonmead at gmail.com> wrote:
> > I guess my question is, how to I type LHS so it is no more specialised as
> > RHS in this case?
>
> I don't believe you can.
>
> >
> > On Sun, Jul 19, 2015 at 6:42 PM, Ivan Lazar Miljenovic
> > <ivan.miljenovic at 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 <clintonmead at 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
> >> > <ivan.miljenovic at 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 <clintonmead at 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
> >> >> > <ivan.miljenovic at gmail.com> wrote:
> >>
> >> >>
> >> >> >>
> >> >> >> On 19 July 2015 at 10:13, Clinton Mead <clintonmead at 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 at gmail.com
> >> >> >> http://IvanMiljenovic.wordpress.com
> >> >> >
> >> >> >
> >> >>
> >> >>
> >> >>
> >> >> --
> >> >> Ivan Lazar Miljenovic
> >> >> Ivan.Miljenovic at gmail.com
> >> >> http://IvanMiljenovic.wordpress.com
> >> >
> >> >
> >>
> >>
> >>
> >> --
> >> Ivan Lazar Miljenovic
> >> Ivan.Miljenovic at gmail.com
> >> http://IvanMiljenovic.wordpress.com
> >
> >
>
>
>
> --
> Ivan Lazar Miljenovic
> Ivan.Miljenovic at gmail.com
> http://IvanMiljenovic.wordpress.com
>
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