Array.prototype.sort()

compareFn Optional

Specifies a function that defines the sort order. If omitted, the array elements are converted to strings, then sorted according to each character's Unicode code point value.

a

The first element for comparison.

b

The second element for comparison.

The reference to the original array, now sorted. Note that the array is sorted in place, and no copy is made.

If compareFn is not supplied, all non-undefined array elements are sorted by converting them to strings and comparing strings in UTF-16 code units order. For example, "banana" comes before "cherry". In a numeric sort, 9 comes before 80, but because numbers are converted to strings, "80" comes before "9" in the Unicode order. All undefined elements are sorted to the end of the array.

If compareFn is supplied, all non-undefined array elements are sorted according to the return value of the compare function (all undefined elements are sorted to the end of the array, with no call to compareFn).

So, the compare function has the following form:

function compareFn(a, b) {
  if (a is less than b by some ordering criterion) {
    return -1;
  }
  if (a is greater than b by the ordering criterion) {
    return 1;
  }
  // a must be equal to b
  return 0;
}

More formally, the comparator is expected to have the following properties, in order to ensure proper sort behavior:

• Pure: The comparator does not mutate the objects being compared or any external state. (This is important because there's no guarantee when and how the comparator will be called, so any particular call should not produce visible effects to the outside.)
• Stable: The comparator returns the same result with the same pair of input.
• Reflexive: compareFn(a, a) === 0.
• Symmetric: compareFn(a, b) and compareFn(b, a) must both be 0 or have opposite signs.
• Transitive: If compareFn(a, b) and compareFn(b, c) are both positive, zero, or negative, then compareFn(a, c) has the same positivity as the previous two.

A comparator conforming to the constraints above will always be able to return all of 1, 0, and -1, or consistently return 0. For example, if a comparator only returns 1 and 0, or only returns 0 and -1, it will not be able to sort reliably because symmetry is broken. A comparator that always returns 0 will cause the array to not be changed at all, but is reliable nonetheless.

The default lexicographic comparator satisfies all constraints above.

To compare numbers instead of strings, the compare function can subtract b from a. The following function will sort the array in ascending order (if it doesn't contain Infinity and NaN):

function compareNumbers(a, b) {
  return a - b;
}

The sort method can be conveniently used with function expressions or arrow functions.

const numbers = [4, 2, 5, 1, 3];
numbers.sort(function (a, b) {
  return a - b;
});
console.log(numbers);
// [1, 2, 3, 4, 5]

// OR

const numbers2 = [4, 2, 5, 1, 3];
numbers2.sort((a, b) => a - b);
console.log(numbers2);
// [1, 2, 3, 4, 5]

Arrays of objects can be sorted by comparing the value of one of their properties.

const items = [
  { name: 'Edward', value: 21 },
  { name: 'Sharpe', value: 37 },
  { name: 'And', value: 45 },
  { name: 'The', value: -12 },
  { name: 'Magnetic', value: 13 },
  { name: 'Zeros', value: 37 }
];

// sort by value
items.sort((a, b) => a.value - b.value);

// sort by name
items.sort((a, b) => {
  const nameA = a.name.toUpperCase(); // ignore upper and lowercase
  const nameB = b.name.toUpperCase(); // ignore upper and lowercase
  if (nameA < nameB) {
    return -1;
  }
  if (nameA > nameB) {
    return 1;
  }

  // names must be equal
  return 0;
});

The following example creates four arrays and displays the original array, then the sorted arrays. The numeric arrays are sorted without a compare function, then sorted using one.

const stringArray = ['Blue', 'Humpback', 'Beluga'];
const numberArray = [40, 1, 5, 200];
const numericStringArray = ['80', '9', '700'];
const mixedNumericArray = ['80', '9', '700', 40, 1, 5, 200];

function compareNumbers(a, b) {
  return a - b;
}

stringArray.join(); // 'Blue,Humpback,Beluga'
stringArray.sort(); // ['Beluga', 'Blue', 'Humpback']

numberArray.join(); // '40,1,5,200'
numberArray.sort(); // [1, 200, 40, 5]
numberArray.sort(compareNumbers); // [1, 5, 40, 200]

numericStringArray.join(); // '80,9,700'
numericStringArray.sort(); // ['700', '80', '9']
numericStringArray.sort(compareNumbers); // ['9', '80', '700']

mixedNumericArray.join(); // '80,9,700,40,1,5,200'
mixedNumericArray.sort(); // [1, 200, 40, 5, '700', '80', '9']
mixedNumericArray.sort(compareNumbers); // [1, 5, '9', 40, '80', 200, '700']

For sorting strings with non-ASCII characters, i.e. strings with accented characters (e, é, è, a, ä, etc.), strings from languages other than English, use String.localeCompare. This function can compare those characters so they appear in the right order.

const items = ['réservé', 'premier', 'communiqué', 'café', 'adieu', 'éclair'];
items.sort((a, b) => a.localeCompare(b));

// items is ['adieu', 'café', 'communiqué', 'éclair', 'premier', 'réservé']

The compareFn can be invoked multiple times per element within the array. Depending on the compareFn's nature, this may yield a high overhead. The more work a compareFn does and the more elements there are to sort, it may be more efficient to use map() for sorting. The idea is to traverse the array once to extract the actual values used for sorting into a temporary array, sort the temporary array, and then traverse the temporary array to achieve the right order.

// the array to be sorted
const data = ['delta', 'alpha', 'charlie', 'bravo'];

// temporary array holds objects with position and sort-value
const mapped = data.map((v, i) => {
  return { i, value: someSlowOperation(v) };
})

// sorting the mapped array containing the reduced values
mapped.sort((a, b) => {
  if (a.value > b.value) {
    return 1;
  }
  if (a.value < b.value) {
    return -1;
  }
  return 0;
});

const result = mapped.map((v) => data[v.i]);

There is an open source library available called mapsort which applies this approach.

The sort() method returns a reference to the original array, so mutating the returned array will mutate the original array as well.

const numbers = [3, 1, 4, 1, 5];
const sorted = numbers.sort((a, b) => a - b);
// numbers and sorted are both [1, 1, 3, 4, 5]
sorted[0] = 10;
console.log(numbers[0]); // 10

In case you want sort() to not mutate the original array, but return a shallow-copied array like other array methods (e.g. map()) do, you can do a shallow copy before calling sort(), using the spread syntax or Array.from().

const numbers = [3, 1, 4, 1, 5];
// [...numbers] creates a shallow copy, so sort() does not mutate the original
const sorted = [...numbers].sort((a, b) => a - b);
sorted[0] = 10;
console.log(numbers[0]); // 3

Since version 10 (or EcmaScript 2019), the specification dictates that Array.prototype.sort is stable.

For example, say you had a list of students alongside their grades. Note that the list of students is already pre-sorted by name in alphabetical order:

const students = [
  { name: "Alex",   grade: 15 },
  { name: "Devlin", grade: 15 },
  { name: "Eagle",  grade: 13 },
  { name: "Sam",    grade: 14 }
];

After sorting this array by grade in ascending order:

students.sort((firstItem, secondItem) => firstItem.grade - secondItem.grade);

The students variable will then have the following value:

[
  { name: "Eagle",  grade: 13 },
  { name: "Sam",    grade: 14 },
  { name: "Alex",   grade: 15 }, // original maintained for similar grade (stable sorting)
  { name: "Devlin", grade: 15 }  // original maintained for similar grade (stable sorting)
];

It's important to note that students that have the same grade (for example, Alex and Devlin), will remain in the same order as before calling the sort. This is what a stable sorting algorithm guarantees.

Before version 10 (or EcmaScript 2019), sort stability was not guaranteed, meaning that you could end up with the following:

[
  { name: "Eagle",  grade: 13 },
  { name: "Sam",    grade: 14 },
  { name: "Devlin", grade: 15 }, // original order not maintained
  { name: "Alex",   grade: 15 }  // original order not maintained
];

If a comparing function does not satisfy all of purity, stability, reflexivity, symmetry, and transitivity rules, as explained in the description, the program's behavior is not well-defined.

For example, consider this code:

const arr = [3, 1, 4, 1, 5, 9];
const compareFn = (a, b) => a > b ? 1 : 0;
arr.sort(compareFn);

The compareFn function here is not well-formed, because it does not satisfy symmetry: if a > b, it returns 1; but by swapping a and b, it returns 0 instead of a negative value. Therefore, the resulting array will be different across engines. For example, V8 (used by Chrome, Node.js, etc.) and JavaScriptCore (used by Safari) would not sort the array at all and return [3, 1, 4, 1, 5, 9], while SpiderMonkey (used by Firefox) will return the array sorted ascendingly, as [1, 1, 3, 4, 5, 9].

However, if the compareFn function is changed slightly so that it returns -1 or 0:

const arr = [3, 1, 4, 1, 5, 9];
const compareFn = (a, b) => a > b ? -1 : 0;
arr.sort(compareFn);

Then V8 and JavaScriptCore sorts it descendingly, as [9, 5, 4, 3, 1, 1], while SpiderMonkey returns it as-is: [3, 1, 4, 1, 5, 9].

Due to this implementation inconsistency, you are always advised to make your comparator well-formed by following the five constraints.

• Polyfill of Array.prototype.sort with modern behavior like stable sort in core-js
• Array.prototype.reverse()
• String.prototype.localeCompare()
• About the stability of the algorithm used by V8 engine
• V8 sort stability
• Mathias Bynens' sort stability demo