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require-array-sort-compare

Require Array#sort calls to always provide a compareFunction.

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 language specification thusly:

note

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.

https://www.ecma-international.org/ecma-262/9.0/#sec-sortcompare

Attributes

  • Included in configs
    • ✅ Recommended
    • 🔒 Strict
  • Fixable
    • 🔧 Automated Fixer
    • 🛠 Suggestion Fixer
  • 💭 Requires type information

Rule Details

This rule aims to ensure all calls of the native Array#sort method provide a compareFunction, while ignoring calls to user-defined sort methods.

Examples of code for this rule:

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

array.sort();

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

Options

The rule accepts an options object with the following properties:

type Options = {
/**
* If true, an array which all elements are string is ignored.
*/
ignoreStringArrays?: boolean;
};

const defaults = {
ignoreStringArrays: false,
};

ignoreStringArrays

Examples of code for this rule with { ignoreStringArrays: true }:

const one = 1;
const two = 2;
const three = 3;
[one, two, three].sort();

When Not To Use It

If you understand the language specification enough, you can turn this rule off safely.