Why does the Java regular expression "|" find a matching substring for any input string?

Question

I am trying to understand why a regular expression ending with "|" (or simply "|" itself) will find a matching substring with start index 0 and end "offset after the last character matched (as per JavaDoc for Matcher)" 0.

The following code demonstrates this:

public static void main(String[] args) {

    String regExp = "|";
    String toMatch = "A";
    Matcher m = Pattern.compile(regExp).matcher(toMatch);
    System.out.println("ReqExp: " + regExp + 
            " found " + toMatch + "(" + m.find() + ") " +  
            " start: " + m.start() +
            " end: " + m.end());

}

Output is:

ReqExp: | found A(true)  start: 0 end: 0

I'm confused by the fact that it is even a valid regular expression. And further confused by the fact that start and end are both 0.

Hoping someone can explain this to me.

user94559 · Accepted Answer · 2017-09-26 00:18:38Z

4

The pipe in a regular expression means "or." So your regular expression is basically "(empty string) or (empty string)". It successfully finds an empty string at the beginning of the string, and an empty string has a length of 0.

edited Sep 26, 2017 at 0:18

answered Sep 26, 2017 at 0:10

user94559

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1 Comment

Pshemo Over a year ago

If someone is confused with (nothing) then maybe it will be clearer if we describe it as empty string "" (one with length 0). In other words we can assume that "|" regex is same as ""+"|"+"" so empty string OR empty string. Since every string can be represented as someString <==> ""+someString we can assume that each string starts (and also ends) with empty string at position 0.

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