How do I split a string into an array of characters? [duplicate]

How do I split a string into an array of characters? [duplicate]

This question already has an answer here:

How do you get a string to a character array in JavaScript?

10 answers

var s = “overpopulation”;
var ar = [];
ar = s.split();

I want to string.split a word into array of characters.
The above code doesn’t seem to work – it returns “overpopulation” as Object..
How do i split it into array of characters, if original string doesn’t contain commas and whitespace?


Solution 1:

You can split on an empty string:

var chars = "overpopulation".split('');

If you just want to access a string in an array-like fashion, you can do that without split:

var s = "overpopulation";
for (var i = 0; i < s.length; i++) {

You can also access each character with its index using normal array syntax. Note, however, that strings are immutable, which means you can’t set the value of a character using this method, and that it isn’t supported by IE7 (if that still matters to you).

var s = "overpopulation";

console.log(s[3]); // logs 'r'

Solution 2:

Old question but I should warn:

Do NOT use .split('')

You’ll get weird results with non-BMP (non-Basic-Multilingual-Plane) character sets.

Reason is that methods like .split() and .charCodeAt() only respect the characters with a code point below 65536; bec. higher code points are represented by a pair of (lower valued) “surrogate” pseudo-characters.

'???'.length     // —> 6
'???'.split('')  // —> ["�", "�", "�", "�", "�", "�"]

'?'.length      // —> 2
'?'.split('')   // —> ["�", "�"]

Use ES2015 (ES6) features where possible:

Using the spread operator:

let arr = [...str];

Or Array.from

let arr = Array.from(str);

Or split with the new u RegExp flag:

let arr = str.split(/(?!$)/u;


[...'???']        // —> ["?", "?", "?"]
[...'???']     // —> ["?", "?", "?"]

For ES5, options are limited:

I came up with this function that internally uses MDN example to get the correct code point of each character.

function stringToArray() {
  var i = 0,
    arr = [],
  while (!isNaN(codePoint = knownCharCodeAt(str, i))) {
  return arr;

This requires knownCharCodeAt() function and for some browsers; a String.fromCodePoint() polyfill.

if (!String.fromCodePoint) {
// ES6 Unicode Shims 0.1 , © 2012 Steven Levithan , MIT License
    String.fromCodePoint = function fromCodePoint () {
        var chars = [], point, offset, units, i;
        for (i = 0; i < arguments.length; ++i) {
            point = arguments[i];
            offset = point - 0x10000;
            units = point > 0xFFFF ? [0xD800 + (offset >> 10), 0xDC00 + (offset & 0x3FF)] : [point];
            chars.push(String.fromCharCode.apply(null, units));
        return chars.join("");


stringToArray('???')     // —> ["?", "?", "?"]
stringToArray('???')  // —> ["?", "?", "?"]

Note: str[index] (ES5) and str.charAt(index) will also return weird results with non-BMP charsets. e.g. '?'.charAt(0) returns "�".

UPDATE: Read this nice article about JS and unicode.

Solution 3:

It’s as simple as:


The delimiter is an empty string, hence it will break up between each single character.

Solution 4:

The split() method in javascript accepts two parameters: a separator and a limit.
The separator specifies the character to use for splitting the string. If you don’t specify a separator, the entire string is returned, non-separated. But, if you specify the empty string as a separator, the string is split between each character.



will have the effect you seek.

More information here

Solution 5:

You can use the regular expression /(?!$)/:


The negative look-ahead assertion (?!$) will match right in front of every character.

Solution 6:

A string in Javascript is already a character array.

You can simply access any character in the array as you would any other array.

var s = "overpopulation";
alert(s[0]) // alerts o.


As is pointed out in the comments below, the above method for accessing a character in a string is part of ECMAScript 5 which certain browsers may not conform to.

An alternative method you can use is charAt(index).

var s = "overpopulation";
    alert(s.charAt(0)) // alerts o.