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

[Clinton Mead]

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 <
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
>
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