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

[Ivan Lazar Miljenovic]

(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