How do I use fix, and how does it work?

Question

I was a bit confused by the documentation for fix (although I think I understand what it\'s supposed to do now), so I looked at the source code. That left me m

Answer 1

You need a way for the fixpoint to terminate. Expanding your example it's obvious it won't finish:

fix id
--> let x = id x in x
--> id x
--> id (id x)
--> id (id (id x))
--> ...

Here is a real example of me using fix (note I don't use fix often and was probably tired / not worrying about readable code when I wrote this):

(fix (\f h -> if (pred h) then f (mutate h) else h)) q

WTF, you say! Well, yes, but there are a few really useful points here. First of all, your first fix argument should usually be a function which is the 'recurse' case and the second argument is the data on which to act. Here is the same code as a named function:

getQ h
      | pred h = getQ (mutate h)
      | otherwise = h

If you're still confused then perhaps factorial will be an easier example:

fix (\recurse d -> if d > 0 then d * (recurse (d-1)) else 1) 5 -->* 120

Notice the evaluation:

fix (\recurse d -> if d > 0 then d * (recurse (d-1)) else 1) 3 -->
let x = (\recurse d -> if d > 0 then d * (recurse (d-1)) else 1) x in x 3 -->
let x = ... in (\recurse d -> if d > 0 then d * (recurse (d-1)) else 1) x 3 -->
let x = ... in (\d -> if d > 0 then d * (x (d-1)) else 1) 3

Oh, did you just see that? That x became a function inside our then branch.

let x = ... in if 3 > 0 then 3 * (x (3 - 1)) else 1) -->
let x = ... in 3 * x 2 -->
let x = ... in 3 * (\recurse d -> if d > 0 then d * (recurse (d-1)) else 1) x 2 -->

In the above you need to remember x = f x, hence the two arguments of x 2 at the end instead of just 2.

let x = ... in 3 * (\d -> if d > 0 then d * (x (d-1)) else 1) 2 -->

And I'll stop here!