[RodgerTheGreat]

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]

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

[RodgerTheGreat]

Amending a slice would amend only the slice:

    l_inner:l[1][1]
    # {"age":3}
    l_inner.age:9
    # {"age":9}
    l_inner
    # {"age":9}
    l
    # (1,(2,{"age":3}),4)

If an amending expression isn't "rooted" in a variable binding, it also returns the entire new structure:

    (1,(list 2,list ().age:5),4)[1][1].age:99
    # (1,(2,{"age":99}),4)