lifting functions to tuples?

[sebc at macs.hw.ac.uk]

Again, I think what you propose is different from what was asked.

> On 2003-11-18 at 10:46EST "Abraham Egnor" wrote:
> > The classic way to write a lift function for tuples is, of course:
> >
> > liftTup f (a, b) = (f a, f b)
> >
> > which has a type of (a -> b) -> (a, a) -> (b, b).  I've been wondering if
> > it would be possible to write a function that doesn't require the types in
> > the tuple to be the same, just that the types in the second tuple are the
> > result of applying the type transformation implied in the function to be
> > lifted to the types in the first tuple.

On Wed, Nov 19, 2003 at 05:25:19PM -0800, oleg at pobox.com wrote:
> 
> *Main> :t liftp
> forall a2 a a3 a1.
> (Typeable a, Typeable a1, Typeable a2, Typeable a3) =>
> (Dynamic -> Dynamic) -> (a1, a3) -> (a, a2)

Here there is no visible relation between the types in the first tuple
and the types in the second tuple.  The ``type transformation implied
in the function to be lifted'' is not reflected in the type of liftp.

I think an example of what Abraham Egnor asked for is:

> f1 :: a -> Maybe a
> f1 x = Just x

with desired type (Maybe Int, Maybe Bool) for the expression
liftTup f1 (1, True).

Yet another example is

> data T a = T (a, a -> T a)
> 
> f2 :: T a -> T a
> f2 (T (repr, m)) = m repr

with desired type (T a, T b) -> (T a, T b) for liftTup f2, so that
one can for example have

> o1 :: Int -> T Int
> o1 x = T (x, \ x -> o1 (x * 2))
> 
> o2 :: Float -> T Float
> o2 x = T (x, \ x -> o2 (x * 2.0))
> 
> v = liftTup f2 (o1 1, o2 1.0)

One could expect v to have type (T Int, T Float), but I don't think
the type system of Haskell can handle these simple, uncontrived
examples.  I don't intend this as a criticism, merely a fact.

Of course, one can always defeat the static type system when it gets
in the way, using Dynamics, as you demonstrated.

-- 
Sebastien

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