3 ms·

Can you give an example of what you mean?

• eru 13y ago
  E.g. the type of map is map :: forall a b. (a -> b) -> [a] -> [b] The important part here is the `forall'. This means, that Haskell guarantees that essentially the map function can not look at the elements of your list. The only way for map to produce a value of type `b' is to apply the given call-back on an `a'. So you know that `map' can not eg add up your `a's. Compare C++'s templates and overloading, where more function definitions with more specific types have precedence over more general ones. See the first answer of https://stackoverflow.com/questions/18180428/is-parametric-polymorphism-the-same-as-dispatching-on-arity https://stackoverflow.com/questions/18180428/is-parametric-p... for some discussion.
  • deleted 13y ago
    [deleted]
  • detrino 13y ago
    This doesn't appear to have anything to do with type classes or overloading, but rather with (unconstrained) parametric polymorphism. Such a function is easy to write in C++, but the function prototype can not provide a strong guarantee that your b value will only be produced by your callback.
    • sixbrx 13y ago
      > but the function prototype can not provide a strong guarantee It doesn't provide any such guarantee, and I think that was the whole point.
    • dllthomas 13y ago
      There are limits placed on you by parametric polymorphism. Adding typeclass constraints in Haskell is how you weaken those limits in a controlled way. So I think it has quite a bit to do with typeclasses. I agree "overloading" might not be the correct term to use here, although it is itself a bit... overloaded.