sort method

void sort ([int compare(E a, E b) ])
override

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

The compare function must act as a Comparator.

List<String> numbers = ['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.

List<int> nums = [13, 2, -11];
nums.sort();
print(nums);  // [-11, 2, 13]

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:

List<String> numbers = ['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

void sort([int compare(E a, E b)]) {
  // Note: arr.sort(null) is a type error in FF
  callMethod('sort', compare == null ? [] : [compare]);
}