There are a few different ways to cook this up. Here's one:
{-# LANGUAGE TemplateHaskell #-}
import Control.Lens
data Cat = Cat { _age :: Int }
deriving Show
makeLenses ''Cat
data Item
= I Int
| L [Item]
| C Cat
deriving Show
makePrisms ''Item
cat :: Cat
cat = Cat 3
l :: [Item]
l = [I 1, L [I 2, C cat], I 4]
l' :: [Item]
l' = set (ix 1 . _L . ix 1 . _C . age) 9 l
ghci> l'
[I 1,L [I 2,C (Cat {_age = 9})],I 4]
In Lil[0], this is how ordinary assignment syntax works. Implicitly defining a dictionary stored in a variable named "cat" with a field "age":
cat.age:3
# {"age":3}
Defining "l" as in the example in the article. We need the "list" operator to enlist nested values so that the "," operator doesn't concatenate them into a flat list:
l:1,(list 2,list cat),4
# (1,(2,{"age":3}),4)
Updating the "age" field in the nested dictionary. Lil's basic datatypes are immutable, so "l" is rebound to a new list containing a new dictionary, leaving any previous references undisturbed:
The website asks what they do in Haskell. The answer is property modification and reading, as well as very powerful traversal constructs, use lenses (https://hackage.haskell.org/package/lens , tutorial at https://hackage.haskell.org/package/lens-tutorial-1.0.5/docs...).
What would be the equivalent to this in Haskell (with or without lens):
That would give us l equal to:In Haskell this list is not well-typed
There are a few different ways to cook this up. Here's one:In Lil[0], this is how ordinary assignment syntax works. Implicitly defining a dictionary stored in a variable named "cat" with a field "age":
Defining "l" as in the example in the article. We need the "list" operator to enlist nested values so that the "," operator doesn't concatenate them into a flat list: Updating the "age" field in the nested dictionary. Lil's basic datatypes are immutable, so "l" is rebound to a new list containing a new dictionary, leaving any previous references undisturbed: There's no special "infix" promotion syntax, so that last example would be: [0] http://beyondloom.com/tools/trylil.htmlThis is surprising to me:
How come it doesn't return just: Or is there something totally different going on with references here? As in, how is this different to:Amending a slice would amend only the slice:
If an amending expression isn't "rooted" in a variable binding, it also returns the entire new structure:Yeah this looks like lenses at first glance
It seems like this is proposing syntactic sugar to make mutating and non-mutating operations be on equal footing.
> The more interesting example is reassigning the deeply nested l to make the cat inside older, without mutating the original cat
Isn't that mutating l, though? If you're concerned about mutating cat, shouldn't you be concerned about mutating l?
It doesn't mutate l exactly, it makes a new list slightly different from the original one and assigns it to l.
That means if someone has a reference to the original l, they do not see the change (because l is immutable. Both of them).
I think I am misunderstanding the behavior of the alt keyword
q has the concept of amend, which is similar: https://code.kx.com/q4m3/6_Functions/#683-general-form-of-fu...
It's quite handy, though the syntax for it is rather clunky compared to the rest of the language in my opinion.