Build and update a list

A list contains values of one type. Use list[T] in signatures and annotations. An empty list needs an annotation or a typed constructor such as list[i64]().

def main() -> Result[(), AllocError | IoError]:
    mut original: list[i64] = [10, 20]?
    mut changed = copy(original)?
    changed.append(30)?
    changed[0] = 99
    print(original)?
    print(changed)?
    print(changed[-1])?
    print(len(changed))?
    Ok(())

[10, 20]
[99, 20, 30]
30
3

copy(original)? makes an independent list. Updates happen in place and require a mut owner. append grows the list; indexed assignment replaces an existing element. Negative indices count from the end, and an invalid index is a runtime error.

List construction, copying, and growth return allocation results, handled here with ?.

Collections can nest. Update an inner list directly through its indices:

def main() -> Result[(), AllocError | IoError | OverflowError[i64]]:
    mut rows = [[1, 2]?, [3]?]?
    rows[0][-1] += 7
    print(rows)?
    Ok(())

[[1, 9], [3]]

Nested assignment does not copy or allocate the containing lists. Numeric elements support augmented assignment such as += too. Each target index runs once before the right side, which may read the list but cannot change or resize it during the update. A failed right-side result leaves the element unchanged. Indices can be expressions, including elements of another list:

def main() -> Result[(), AllocError | IoError]:
    mut values = [30, 10, 20]?
    order = [1, 2, 0]?
    print(values[order[0]])?
    values[order[2]] = 31
    print(values)?
    Ok(())

10
[31, 10, 20]