Reference · Lists

push

push(list | null, any=) -> list

What it does

A new list with one item added at the end. Values never change in Quillo, so the list given stays as it was and the longer list is the result: assign it back (xs = xs | push(v)) to keep it. null counts as an empty list.

Examples

print(push([1, 2], 3))let todos = ["milk"]todos = todos | push("eggs")print(todos)print(push([], { id: 1 }))

[1, 2, 3]["milk", "eggs"][{ id: 1 }]

Edge cases

• The original is unchanged: after let b = push(a, 3), a is as it was.
• push(null, x) gives [x], so a list that may still be null needs no check.
• Only one item is added, as it is: pushing a list adds it as one nested item. Use + to join two lists.
• The item left out is null: push(xs) adds a null.

let a = [1, 2]let b = push(a, 3)print(a, b)print(push(null, 1), push([1], [2, 3]), [1] + [2, 3])print(push([1]), push([1], null))

[1, 2] [1, 2, 3][1] [1, [2, 3]] [1, 2, 3][1, null] [1, null]

Errors

A first argument that is not a list or null is a runtime error, not an error value: a string is not a list of its characters, and a map is not a list.

print(push("a", "b"))

line 1: `push` expects a list as argument 1, got str

Chaining

In a pipeline the list comes first and the item is the only argument: xs | push(v). Building a list item by item is a reduce whose value so far is a list.

print([] | push(1) | push(2))print(reduce([1, 2, 3], [], (acc, x) => push(acc, x * 2)))let parts = ["a"]for p in ["b", "c"] {    parts = parts | push(p)}print(join(parts, "-"))

[1, 2][2, 4, 6]a-b-c

See also

reduce, set, remove, flatten, join

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