Generalised plusequals

(leontrolski.github.io)

15 points | by leontrolski 14 hours ago ago

10 comments

$flebron 11 hours ago

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...).
[-]
- $leontrolski 3 hours ago
  What would be the equivalent to this in Haskell (with or without lens):
```
    cat = Cat(age=3)
    l = [1, [2, cat], 4]
    alt l[1][1].age.=9
```
  That would give us l equal to:
```
    [1, [2, Cat(age=9)], 4]
```
$RodgerTheGreat 10 hours ago
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:
```
    l[1][1].age:9
    # (1,(2,{"age":9}),4)
    cat
    # {"age":3}
```
There's no special "infix" promotion syntax, so that last example would be:
```
    l:l,5
    # (1,(2,{"age":9}),4,5)
```
[0] http://beyondloom.com/tools/trylil.html
[-]
- $leontrolski 3 hours ago
  This is surprising to me:
```
  l[1][1].age:9
  # (1,(2,{"age":9}),4)
```
  How come it doesn't return just:
```
  {"age":9}
```
  Or is there something totally different going on with references here? As in, how is this different to:
```
  l_inner = l[1][1]
  l_inner.age:9
```
$rokob 10 hours ago

Yeah this looks like lenses at first glance
$hatthew 10 hours ago

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?
[-]
- $two_handfuls 10 hours ago
  
  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).
  [-]
  - $ 6 hours ago
    
    [deleted]
  - $hatthew 6 hours ago
    
    I think I am misunderstanding the behavior of the alt keyword
$beaumayns 10 hours ago

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.