Question

In arrow do notation, you can use the rec keyword to write recursive definitions. So for example:

rec
    name <- function -< input
    input <- otherFunction -< name

How can this ever evaluate? It seems like it would just go into an infinite loop or something. I know it evaluates to the loop arrow combinator, but I don’t understand how that works either.

EDIT: that powers example is really helpful. How would you write that with do notation, though? I assume you would need to use rec.

Answer 1

This bit of magic works due to haskells laziness. As you might know, Haskell evaluates a value when needed, not when defined. Thus, this works if you don’t need the value fed in directly, or maybe later.

rec is implemented using the loop function of ArrowLoop. It is defined as followed:

class Arrow a => ArrowLoop a where
        loop :: a (b,d) (c,d) -> a b c

instance ArrowLoop (->) where
        loop f b = let (c,d) = f (b,d) in c

You can see: The output is just fed back as the input. It will be calculated just once, because Haskell will only evaluate d when it’s needed.

Here’s an actual example of how to use the loop combinator directly. This function calculates all the powers of it’s argument:

powers = loop $ \(x,l) -> (l,x:map(*x)l)

(You could also write it like this instead: powers x = fix $ (x :) . map (*x))

How does it works? Well, the infinite list of powers is in the l argument. The evaluation looks like this:

powers = loop $ \(x,l) -> (l,x:map(*x)l) ==>
powers b = let (c,d) = (\(x,l) -> (l,x:map(*x)l)) (b,d) in c ==>
powers b = let (c,d) = (d,b:map(*b)d) in d ==> -- Now  we apply 2 as an argument
powers 2 = let (c,d) = (d,2:map(*2)d) in d ==>
         = let (c,(2:d)) = (d,2:map(*2)d) in c ==>
         = let (c,(2:4:d)) = ((2:d),2:map(*2)(2:d)) in c ==>
         = let (c,(2:4:8:d)) = ((2:4:d),2:map(*2)(2:4:d)) in  ==> -- and so on