Why do javascript string split method support an array as separator?

Question

From the MDN the split method separator is treated as a String or a RegExp. But 'asd'.split(['s']) correctly returns ['a', 'd'].

A lot of JS functions will try to coerce their argument(s) to the required type(s). ['s'].toString() gives 's'. — nnnnnn
– nnnnnn, Commented Jul 20, 2016 at 9:13
MDN says the separator is treated as a string. So this probably works because String(['s']) === 's'. — str
– str, Commented Jul 20, 2016 at 9:13
"If separator is a RegExp object (its [[Class]] is "RegExp"), let R = separator; otherwise let R = ToString(separator)." — evolutionxbox
– evolutionxbox, Commented Jul 20, 2016 at 9:15
the coercion rules as said by nnnnnn in fact "111a,s222".split(['a','s']) splits since ""+['a','s'] == 'a,s' — fedeghe
– fedeghe, Commented Jul 20, 2016 at 9:16

James Thorpe · Accepted Answer · 2016-07-20 09:17:27Z

4

The specification for split includes the following step:

Let R be ToString(separator).

So the array is coerced to a string, and the string representation of ['s'] is s.

It's worth noting that if you pass a regex to split, it doesn't get this far, as the regex object has an internal @@split method, which is used in step 3:

If separator is neither undefined nor null, then

a. Let splitter be GetMethod(separator, @@split).

b. ReturnIfAbrupt(splitter).

c. If splitter is not undefined , then

i. Return Call(splitter, separator, «‍O, limit»).

edited Jul 20, 2016 at 9:17

answered Jul 20, 2016 at 9:13

James Thorpe

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1 Comment

Mohit Bhardwaj Over a year ago

Yes, you're right, I just checked it with an example console.log('as,df'.split(['s', 'd']) );

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