Got an idea, is it the right path? Are there bad consequences in using the extensions that are necessaries to make the code work?
class Mdk binary a | binary -> a where
mdk :: a -> b -> binary
instance Mdk (a,b) a where
mdk = ... :: a -> b -> (a,b)
instance Mdk (b, a) a where
mdk = ... :: a -> b -> (b, a)
then, however, even if it works, mdk will never be part of the other class i was working on, and at that point i'm a bit lost on the consequences of this...
Le samedi 4 juin 2016, Silent Leaf <silent.leaf0@gmail.com> a écrit :
> I got a function, named mdk (don't ask). It happens it's a method of a class, as follows:
>
> class C a where
> mdk :: C b => a -> b -> (a, b)
>
> yet i got another function, called so far mdk'
> mdk' :: C b => b -> a -> (a, b)
>
>
> Say there are the following instances:
> instance C X where
> mdk x b = ... :: C b => (X, b)
> mdk' b x = ... :: C b => (X, b)
> instance C Y where
> mdk y b = ... :: C b => (Y, b)
> mdk' b y = ... :: C b => (Y, b)
>
> if you get what's happening, mdk and mdk' are basically the same method, save for the final type-result, which is reversed, without data loss mind you (aka (A, B) and (B, A) could be the same type, if only i knew of a way to make a type in haskell without forcing the order (a kind of set or something).
> so, there are four cases:
> input have types, in this order, X and Y, and expected output is (X, Y): then mdk from X's instance of C is called/must be used;
> input have types Y and X, output (X, Y): X's mdk'
> input X and Y, output (Y, X): Y's mdk
> input Y and X, output (Y, X): Y's mdk'
>
> it might seem confusing, but all amounts to the issue that even though (Y, X) is strictly equivalent, in my program, to (X, Y), in principle haskell differentiate the types.
> thus i have to write two versions of mdk for each type, even though the methods "X's mdk" and "Y's mdk'" are identical, and same for the two other methods.
> indeed:
> mdk :: C b => X -> b -> (X, b)
> mdk' :: C b => b -> Y -> (Y, b)
> are identical signatures, if in the first one you replace `b` with Y, and in the latter you replace `b` with X, and ofc if you consider (X,Y) == (Y,X).
>
> only one of those two methods should be enough to handle the job, and be chosen by the compiler on the sole value of the type of the first variable; but the fact both methods return tuples of "mirrored" types, crushes that. if i only write `mdk` instances, as soon as the compiler will meet this following signature:
> mdk :: a -> b -> (b, a)
> it will crash an exception, because mdk's original signature has for output value a tuple whose first type should here be `a`, because it's meant to be the type of the first argument, not the second.
>
> thus, at last my question: can i tell the compiler to consider (b, a) and (a, b) as identical types (don't worry it's not really a tuple in my program, but it's equivalent)?
> if not, can i make an overloading of mdk so it accepts both a->b->(a,b) and a->b->(b,a)?
>
> hope i didn't lose anyone. if so, do tell me, i'll try to clarify.
>
> thanks in advance of the time spent trying to understand my problem!