5 ms·

How does that work in Haskell? Is the user of a semigroup somehow guaranteed that its operation is associative?

• wereHamster 10y ago
  The person who implements Semigroup for a data type must ensure that the laws hold.
  • deleted 10y ago
    [deleted]
  • haimez 10y ago
    Which is where property based testing really shines. Still requires the implementer to be a good citizen and write those tests though.
    • kristianp 10y ago
      AKA QuickCheck. https://en.wikipedia.org/wiki/QuickCheck https://en.wikipedia.org/wiki/QuickCheck
• tikhonj 10y ago
  No. Guaranteeing typeclass laws like associativity is hard. People have looked into various ways to do that, but haven't found anything completely satisfying. It's either hard for the programmer (dependent typing) or hard to implement (automatic theorem proving). I hope that'll change, but for now Haskell does not have any facilities for specifying or enforcing typeclass laws. As it is, it works more or less exactly like interfaces from other languages. When you implement Comparable in Java you implicitly expect certain laws (ie that it's a valid ordering), but there is also no way to specify or verify them.
  • Bromskloss 10y ago
    Clueless suggestion: What about the programmer (who implements the semigroup) also providing a proof (that can be automatically checked) that it satisfies associativity? Is that method perhaps already covered by one of the examples you mentioned?
    • reuben364 10y ago
      That is the dependent typed aspect mentioned. This is actually done in Idris with VerifiedSemigroup and other verified instances.
      • Bromskloss 10y ago
        OK. I'm afraid I don't know enough to make the connection. Actually, I never understood the point of dependent types. They seem to do away with the neat order that types introduce. ¯\_(ツ)_/¯