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Thanks for your excellent questions. They're really helping me fuzz through my approach - I wish I had people I could do this with more often, haha. :P Anyway, something I realized after I posted that question is that assignment doesn't have to force a copy; for instance, this could just take a CoW slice reference to the original list: [head & tail] = &rem'
The only requirement is that the value returned from a function is an owned value, not CoW. To avoid talking myself into circles, I'll show that `append` still works under this rule. It just moves the copy to the point where we construct the new list, rather that at the point where we de-structure it: loop [head' & acc'] tail'
So in `[head' & acc']`.For the code sample you provided, a reference would be passed down, rather than a copy of the entire list, so just the first two elements would be copied. Do you have any further questions regarding this? I'd like to be able to prove the soundness of this system, but I'm not sure going back and forth on example after example is the best strategy in the long run. > In a functional language, insertion into a binary tree only copies nodes along one path in the tree. Would your language copy the entire tree instead, at potentially exponentially higher cost? Most functional languages use persistent data-types, because all data in those languages are immutable. I haven't thought about how they'd work into Passerine, but if you mutate data that will be used later, a copy of that data will be made. So yes, mutating a tree makes a copy of it if the original tree is used later. This is less of a problem then you'd think. It's common to write programs as series of transformations on data, so note that if you do something like this on a big million-element list
children
.map { child -> (child.name, child.age, child.hobby) }
.filter { (_, age, _) -> age > 10 }
.map { (name, _, hobby) -> (first_name name, hobby) }
.filter { (_, hobby) -> is_fun hobby }
.fold "" { (string, (name, hobby)) -> string + name + " enjoys " + hobby + "\n" }
.print No copies will be made, because each step is the last usage of the data-structure produced in the previous step. > (I'm also slightly confused by your definition of append -- the accumulator seems to be a list, but then the initial call of loop would need to use [value] instead of value, no?) Once again, correct. > Regarding "tying references to the stack" I had another question here: https://news.ycombinator.com/item?id=26228961 I wrote a long response, to this, but it ended up being a couple thousand words, so I'm trying to trim it down. (I went off the deep end, with stack diagrams, explaining how cycles get created, etc, etc.) TL;DR for now is that you can't have a list of references, each object is only mutably 'owned' by one other. Once I trim down the response, I'll post it - should be sometime later today. |