# sort method

1. @override

void sort([

1. int compare(
   1. E a,
   2. E b
   )?

])
override

Sorts this list according to the order specified by the `compare` function.

The `compare` function must act as a Comparator.

    final numbers = <String>['two', 'three', 'four'];
    // Sort from shortest to longest.
    numbers.sort((a, b) => a.length.compareTo(b.length));
    print(numbers); // [two, four, three]

The default List implementations use Comparable.compare if `compare` is omitted.

    final numbers = <int>[13, 2, -11, 0];
    numbers.sort();
    print(numbers); // [-11, 0, 2, 13]

In that case, the elements of the list must be Comparable to each other.

A Comparator may compare objects as equal (return zero), even if they are distinct objects. The sort function is not guaranteed to be stable, so distinct objects that compare as equal may occur in any order in the result:

    final numbers = <String>['one', 'two', 'three', 'four'];
    numbers.sort((a, b) => a.length.compareTo(b.length));
    print(numbers); // [one, two, four, three] OR [two, one, four, three]

## Implementation

    @override
    void sort([int Function(E a, E b)? compare]) {
      // This is highly inefficient for a linked list.
      // Convert to list, sort, then rebuild.
      final List<E> temp = toList();
      temp.sort(compare);
      clear();
      addAll(temp);
    }