Generalised plusequals

(leontrolski.github.io)

15 points by leontrolski 15 hours ago|9 comments

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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...).

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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]

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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

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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

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rokob 10 hours ago

Yeah this looks like lenses at first glance

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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?

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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).

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hatthew 6 hours ago

I think I am misunderstanding the behavior of the alt keyword

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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.