regex - How to avoid Blank space insertion at the beginning of the String in Java while using Split() function -


here code snippet.

public static void main(string...strings){         string s="google";         string[] s1=s.split("");         system.out.println(s1[0]);         system.out.println(s1.length);     }

i'm trying split string s around each character. problem while splitting blank space getting introduced @ begining of splitted array. giving me output length 7 instead of 6. , since can't have trailing spaces split here getting passed split("", 0), has @ beginning test printed s1[0] , giving blank space indeed.

my question how avoid such problem. need use split.

what happening here.?

ideone link

the method string.split() behaves differently in java 7 (used ideone) , java 8 (probably used of commenters , answerers). in java 8 javadoc documentation string.split(), following written:

when there positive-width match @ beginning of string empty leading substring included @ beginning of resulting array. zero-width match @ beginning never produces such empty leading substring.

so, according java 8, "google".split("") equal ["g","o","o","g","l","e"].

this remark not present in java 7 documentation, , indeed, in java 7 seems "google".split("") equal ["", "g","o","o","g","l","e"].

(both in java 7 , in java 8 empty strings @ end of array removed.)

the best solution seems to add code ignore first string if empty. (note split("\\b") solution suggested in answer yields unexpected results when input consists of more 1 word.)


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