4 ms·

I'm not proficient in Haskell, but how would the compiler be able to statically type that expression?

• yorwba 8y ago
  Prelude> y f = f (y f) Prelude> :type y y :: (t -> t) -> t Because f is applied to an argument, it must be a function of generic type t -> u. Because y is applied to f, it must be a function of type (t -> u) -> v. Then the type of y f is v. Because f is applied to y f, t = v. Then the type of f (y f) is u. Because y f is defined as f (y f), the two expressions need to have the same type, t = u. Therefore f is of type t -> t and y is of type (t -> t) -> t.
  • user2994cb 8y ago
    It's true that the original non-recursive definition of Y doesn't typecheck in Haskell: Prelude> \f->(\x->f(x x))(\x->f(x x)) <interactive>:5:14: Occurs check: cannot construct the infinite type: r0 ~ r0 -> t ...