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

Tony Morris tonymorris at gmail.com
Sun Jul 19 08:52:58 UTC 2015


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
> <ivan.miljenovic at gmail.com <mailto: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
>     <mailto: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 <mailto: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 <mailto: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 <mailto:ivan.miljenovic at gmail.com>>
>     wrote:
>     >>
>     >> >>
>     >> >> On 19 July 2015 at 10:13, Clinton Mead <clintonmead at gmail.com
>     <mailto: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 <mailto:Ivan.Miljenovic at gmail.com>
>     >> >> http://IvanMiljenovic.wordpress.com
>     >> >
>     >> >
>     >>
>     >>
>     >>
>     >> --
>     >> Ivan Lazar Miljenovic
>     >> Ivan.Miljenovic at gmail.com <mailto:Ivan.Miljenovic at gmail.com>
>     >> http://IvanMiljenovic.wordpress.com
>     >
>     >
> 
> 
> 
>     --
>     Ivan Lazar Miljenovic
>     Ivan.Miljenovic at gmail.com <mailto:Ivan.Miljenovic at gmail.com>
>     http://IvanMiljenovic.wordpress.com
> 
> 
> 
> 
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