Removing repetitions from multiple instance declarations

Question

I've currently got code following this pattern:

f1' x1 x2 .. xm1 = ...
f2' x1 x2 .. xm2 = ...
.
.
,
fn' x1 x2 .. xmn = ...

instance C D1 where
  f1 = f1'
  f2 = f2'
  .
  .
  .
  fn = fn'

instance C D2 where
  f1 = f1'
  f2 = f2'
  .
  .
  .
  fn = fn'

.
.
.

instance C DN where
  f1 = f1'
  f2 = f2'
  .
  .
  .
  fn = fn'

Basically, I've got a class that I want to implement for a number of datatypes, but all these implementations are the same. But there's a lot of repeated code with the instance implementations, for m functions and n instances, I have to write roughly O(m*n) lines of code, I'd prefer it to be more like O(m+n).

Is there a good way to remove the repetition here? I imagine Template Haskell would do the trick, but I don't want to reinvent the wheel if something to handle this already exists.

If they are all the same, can't you just provide default implementations? — Zeta, Mar 23 '17 at 06:13
I can't do that sadly, I don't control the class (`C` is actually `Enum` in my case). — Clinton, Mar 23 '17 at 06:36
Instead of the abstract example, could you add two of your types as well as your functions? — Zeta, Mar 23 '17 at 08:36
I don't know if there's something simpler than Template Haskell for this. — chi, Mar 23 '17 at 10:28
Sometimes in cases like this I put all the definitions of each instance on single line, eases the visual pain and shows the repetition more clearly. — luqui, Mar 23 '17 at 16:36

score 0 · Answer 1 · edited May 23 '17 at 12:10

0

You can provide a default implementation for your class. Take the Functor example:

class  Functor f  where
    fmap        :: (a -> b) -> f a -> f b

     -- | Replace all locations in the input with the same value.
     -- The default definition is @'fmap' . 'const'@, but this may be
     -- overridden with a more efficient version.
    (<$)        :: a -> f b -> f a
    (<$)        =  fmap . const

You can also opt for the deriving option which you can found more information in this question: How does deriving work in Haskell? There is some information on how to control deriving.

edited May 23 '17 at 12:10

Community

1
1

answered Mar 23 '17 at 09:40

mathk

7,973
6
45
74

C is not my class only the instances are mine – Clinton Mar 23 '17 at 12:03
is deriving an option ? – mathk Mar 23 '17 at 13:11

score 0 · Answer 2 · answered Mar 23 '17 at 16:21

0

Maybe you could first create another empty type class:

class SomeD

Then make the types D1 ... DN instances of SomeD using a single line for each type:

instance SomeD D1
-- ...
instance SomeD DN

Then make any instance of the SomeD typeclass an instance of C like so:

instance (SomeD t) => C t where
    f1 = f1'
    -- ...
    fn = fn'

(I'm not sure if this works or if it needs some language extensions but unfortunately I can't test it right now.)

answered Mar 23 '17 at 16:21

Mekeor Melire

502
5
15

This does not work and there's not really an extension which allows it to. The only case where this works is if there are no other implementations of `C`. Yes you can get it to compile but when you try to use it everything gets messy. – luqui Mar 23 '17 at 16:33

Removing repetitions from multiple instance declarations

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