Java regexp - adding parenthesis makes it greedy?

Question

public void name() throws Exception {
        Pattern p = Pattern.compile("\\d{1,2}?");
        String input = "09";
        Matcher m = p.matcher(input);
        StringBuffer sb = new StringBuffer();
        while(m.find()) {
            System.out.println("match = "+m.group());
        }
    }

Output of the above method is :

match = 0
match = 9

Now, I am just adding parenthesis to the regexp:

public void name() throws Exception {
        Pattern p = Pattern.compile("(\\d{1,2})?");
        String input = "09";
        Matcher m = p.matcher(input);
        StringBuffer sb = new StringBuffer();
        while(m.find()) {
            System.out.println("match = "+m.group());
        }
    }

And the output becomes:

match = 09
match =

Why do parenthesis make the matching greedy here?
[Edit,added later]Why is empty string not matched in the first case?

M A · Accepted Answer · 2017-01-06 19:59:29Z

6

In order for a quantifier (?, *, + or {n,m}) to be reluctant (non-greedy or lazy), it must by followed by a ?. Therefore, the pattern \\d{1,2}? is reluctant.

On the other hand, (\\d{1,2})? is composed of two levels of greedy quantifiers:

a group (\\d{1,2}) containing a pattern with a greedy quantifier {1,2},
followed by the ? greedy quantifier.

Therefore, (\\d{1,2})? is greedy because the ? does not immediately follow a quantifier (there is a parenthesis in between that closes the group), hence it does not act as the metacharacter for a reluctant regex.

See this page about quantifiers as a reference.

edited Jan 6, 2017 at 19:59

answered Jan 6, 2017 at 19:13

M A

73.2k14 gold badges150 silver badges182 bronze badges

Sign up to request clarification or add additional context in comments.

7 Comments

xenteros Over a year ago

I have already upvoted this answer due to the url, but could you elaborate a bit more on the topic?

simplfuzz Over a year ago

Yes, even I would request you to elaborate further.

simplfuzz Over a year ago

Just edited the question, Why is empty string not matched in the first case?

M A Over a year ago

@dharm0us Exactly. A ? right after a quantifier only makes that quantifier lazy.

xenteros Over a year ago

@dharm0us I believe that AR has elaborated enough now. It would be kind of you to accept the answer :)

|

anubhava · Accepted Answer · 2017-01-06 19:32:20Z

1

Let's start with simple quantifier regex:

\d{1,2}

This is greedy by nature and will match as many characters as possible between min (i.e. 1) and max (i.e 2).

So for our input 09, it will just match 09.

Now let's make it lazy by using:

\d{1,2}?

This will make it as few times as possible. So for the same input it will match 0 first time and 9 second time (since you're using a while loop).

Now 3rd case of:

(\d{1,2})?

It matches 0 or 1 occurrence of greedy \d{1,2}, that means means match \d{1,2} or match nothing.

So for the same input it will match:

09 or
empty string

It is because we are making whole group optional by placing ? outside (...)

answered Jan 6, 2017 at 19:32

anubhava

790k67 gold badges603 silver badges671 bronze badges

Collectives™ on Stack Overflow

Java regexp - adding parenthesis makes it greedy?

2 Answers 2

7 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

7 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related