Generalised plusequals
leontrolski.github.ioThe 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:cat = Cat(age=3) l = [1, [2, cat], 4] alt l[1][1].age.=9[1, [2, Cat(age=9)], 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}
l:1,(list 2,list cat),4
# (1,(2,{"age":3}),4)
l[1][1].age:9
# (1,(2,{"age":9}),4)
cat
# {"age":3}
l:l,5
# (1,(2,{"age":9}),4,5)
This is surprising to me:
How come it doesn't return just:l[1][1].age:9 # (1,(2,{"age":9}),4)
Or is there something totally different going on with references here? As in, how is this different to:{"age":9}l_inner = l[1][1] l_inner.age:9
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.