Hello,
I was writing some Haskell when I stumbled on the following problem:
With the following language extension...
> {-# LANGUAGE RankNTypes #-}
... it's possible to define 'foo' and 'bar' like so:
> foo :: (Num c, Num d) => (forall b. Num b => a -> b) -> a -> (c, d)
> foo f x = (f x, f x)
> bar :: (Read c, Read d) => (forall b. Read b => a -> b) -> a -> (c, d)
> bar f x = (f x, f x)
Which allows us to write:
> testFoo = foo fromInteger 1 :: (Int, Float)
> testBar = bar read "1" :: (Int, Float)
Now I would like to generalise 'foo' and 'bar' to 'bla' so that I can write:
testBla1 = bla fromInteger 1 :: (Int, Float)
testBla2 = bla read "1" :: (Int, Float)
My question is how to define 'bla'.
I can write:
> bla :: (forall b. a -> b) -> a -> (c, d)
> bla f x = (f x, f x)
But then 'testBla1' gives the following expected error:
Could not deduce (Num b) from the context ()
arising from a use of `fromInteger'
Possible fix:
add (Num b) to the context of
the polymorphic type `forall b. a -> b'
In the first argument of `bla', namely `fromInteger'
In the expression: bla fromInteger 1 :: (Int, Float)
In the definition of `testBla1':
testBla1 = bla fromInteger 1 :: (Int, Float)
And 'testBla2' gives a similar error complaining that it can't
deduce (Read b) from the context.
So, somehow I need to quantify over the type class.
bla :: forall cls. (cls c, cls d) => (forall b. cls b => a -> b) -> a -> (c, d)
But this isn't legal.
Is there another way of defining 'bla'?
Thanks,
Bas
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