`Array#sort` must have a compare functionJS-0375

This rule prevents invoking the Array#sort() method without providing a compare argument.

When called without a compare function, Array#sort() converts all non-undefined array elements into strings and then compares said strings based off their UTF-16 code units.

The result is that elements are sorted alphabetically, regardless of their type. When sorting numbers, this results in the classic "10 before 2" order:

[1, 2, 3, 10, 20, 30].sort(); //→ [1, 10, 2, 20, 3, 30]

This also means that Array#sort does not always sort consistently, as elements may have custom #toString implementations that are not deterministic; this trap is noted in the noted in the language specification thusly:

NOTE 2: Method calls performed by the ToString abstract operations in steps 5 and 7 have the potential to cause SortCompare to not behave as a consistent comparison function. ttps://www.ecma-international.org/ecma-262/9.0/#sec-sortcompare

Bad Practice

const array: any[];
const stringArray: string[];

array.sort();

// String arrays should be sorted using `String#localeCompare`.
stringArray.sort();

Recommended

const array: any[];
const userDefinedType: { sort(): void };

array.sort((a, b) => a - b);
array.sort((a, b) => a.localeCompare(b));

userDefinedType.sort();