[kannmig]

Another great exposition on the applicative-order Y combinator can be found at the end of Chapter 9 of The Little Schemer.

It is interesting to contrast with how in Haskell, one can simply define Y as

  Y f = f (Y f)

owing to the fact that it is lazily evaluated.

[evincarofautumn]

Of course, the definition we actually use is this:

    -- (1)
    fix f = let x = f x in x

Because it has better memory usage; the definition you provided is translated to the following STG:

    -- (2)
    fix f = let x = fix f in f x

So whereas #1 is a thunk that refers to itself, #2 is a recursive function, and importantly not tail-recursive.

[snrji]

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

[yorwba]

  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]

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
    ...