[Haskell-cafe] A functor for two Peano systems

[Akhra Gannon]

I think your last guess is correct... what you're trying to build appears
to be a semigroup (or monoid), not a functor.

On Sun, Apr 4, 2021, 10:18 PM Galaxy Being <borgauf at gmail.com> wrote:

> I'm just not understanding the concept of a functor in this context: I
> have this
>
> plus :: Int -> Int -> Int
> plus n m = if (n == 0)
>            then m
>            else sCessor (plus (pCessor n) m)
>                 where sCessor x = x + 1
>                       pCessor x = if (x == 0)
>                                   then error "too small"
>                                   else (x - 1)
>
> and this
>
> data MyNum = MNZero | OneMoreThan MyNum deriving (Show,Eq,Ord)
>
> plus2 :: MyNum -> MyNum -> MyNum
> plus2 n m = if (n == MNZero)
>             then m
>             else sCessor (plus2 (pCessor n) m)
>                  where sCessor x = (OneMoreThan x)
>                        pCessor x = if (x == MNZero)
>                                    then (error "too small")
>                                    else (oneLess x)
>                        oneLess MNZero = MNZero
>                        oneLess (OneMoreThan myn) = myn
>
> It seems there should be just one plus, function that would handle both
> an Int-based Peano and the MyNum-based Peano, not two. But in this
> definition
>
> fmap :: (a -> b) -> f a -> f b
>
> The (a -> b) should be "lifted" over the f a -> f b But I can't conceive
> of how this should all fit together, i.e., to create just one generic plus
> that would handle both the plus :: Int -> Int -> Int and the plus2 ::
> MyNum -> MyNum -> MyNum. Trying to get started, I would assume I need
> some sort of instance Functor MyNum describing, yeah, what? Or am I
> barking up the completely wrong tree here, i.e., this isn't a functor issue?
>
> LB
>
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