Java split string on empty delimiter returns empty string at the beginning?

Question

I want to split a string to single characters. So I do:

"abcd".split("");

But this yields:

["", "a", "b", "c", "d"]

The first empty string is not something I'm used to when doing the same in other languages (e.g. Ruby). What is the logic behind it?

You can use substring(i, i+1) or charAt(i) or toCharArray() to access characters. — khachik
– khachik, Commented Apr 19, 2012 at 17:14
But why is the empty string added only to the beginning of the array and not everywhere? (especially at the end) — Gadi A
– Gadi A, Commented Apr 19, 2012 at 17:15
There is no empty string at the end because "[String#split(String)] works as if by invoking the two-argument split method with the given expression and a limit argument of zero. Trailing empty strings are therefore not included in the resulting array." — Matt Ball
– Matt Ball, Commented Apr 19, 2012 at 17:17

Colin D · Accepted Answer · 2012-04-19 17:40:44Z

4

Why are you using String.split() for this? You might be better served using String.toCharArray().

I know one will return you an array of Strings while the other will give you an array of chars. Since you want each character separately, I am assuming this doesn't matter to your code.

answered Apr 19, 2012 at 17:40

Colin D

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1 Comment

Gadi A Over a year ago

I was more interested in the strange behavior, but you are correct; that is a better way.

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