Regex explaination

[Sudhanshu Mishra]

Hi all,
I am confused regarding a regular expression

(?=^[\w|\s])[\w|\s]

Now what this does is matches the first character for any string.
I am having difficulty understanding the behaviour.
Please help me out.

THanks,
Sudhanshu

[Anuj Sharma R]

Thanks Sudhanshu for an interesting question

Let us breakdown your reg-ex:
(?=^[\w|\s])[\w|\s]

1. (x|y) Find any of the alternatives specified within the brackets
2. ^n Matches any string with n at the beginning of it
3. ?=n Matches any string that is followed by a specific string n.
4. \w Find a word character
5. \s Find a whitespace character

Lets start with 2 and apply to our expression:
^[\w|\s] - This means we need word or a space at the start of the line
Let us apply # 3 to this
?=^[\w|\s] - This means we need to match any string which is followed by word or a space at the start of the line
Applying 1 to this would be:
(?=^[\w|\s]) - Since bracket contains only 1 value as defined by above - so it can be any word or space at the start of the line

Hope this clarifies.

[Rob Spoor]

Anuj Sharma R wrote:1. (x|y) Find any of the alternatives specified within the brackets

True, but the regex contains [x|y]. That's a character class, and | has no special meaning inside character classes. In other words, it means one character matching x, | or y.

3. ?=n Matches any string that is followed by a specific string n.

? has two types of meaning: either a quantifier (zero or one) if it follows another part of a regular expression, or it's part of a lookahead / lookbehind if it follows a (. The latter is the case here, so (?=x) means that x must be matched, but it will not be part of the match itself.

^[\w|\s] - This means we need word or a space at the start of the line

Or a | itself.

This regex is a bit unnecessarily complex. The regex without the positive lookahead says that it must be a word character, whitespace or |. The lookahead itself says that from the start of the match, there must be a word character, | or whitespace but only as the start of the string. The entire regex can be shorted to ^[\w|\s].

[Richard Tookey]

Anuj Sharma R wrote:
Hope this clarifies.

Sorry but it does not clarify because your explanation is wrong. Yes (x|y) denotes alternation but [x|y] is being used and this does not denote alternation but a character set that actually contains a '|' ! This means that if used with matches() the regex will match a single word character, space or '|' character and if used with find() will match anything starting with a word character, space or '|' character.

I suspect that whoever wrote that regex did not have a good understanding of regular expressions since the look forward would seem to be redundant.

edit - too slow.

[Anuj Sharma R]

Thanks Rob and Richard for clarifying further on this. I agree this regex is un-necessarily complex.

[Bin Smith]

Behavior of this regex depends on multiline mode of java.util.regex.Pattern.

So when MULTILINE is used then caret symbol ^ denotes start of each line. When MULTILINE is not used, ^ denotes start of entire input string passed to Pattern.matcher() method.

Now to see this difference in practice put this code into main method :

Pattern pattern = Pattern.compile("(?=^[\\w|\\s])[\\w|\\s]", Pattern.MULTILINE);
		Matcher matcher = pattern.matcher("el\n43");
		// System.out.println("matches="+matcher.matches());
		// Pattern pattern = Pattern.compile("([0-9][a-z]*).*\\1");
		// Matcher matcher = pattern.matcher("2aa2a");
		while (matcher.find()) {
			String group = matcher.group();
			System.out.println("group=" + group);
		}

The result is :

group=e
group=4

Next try to remove Pattern.MULTILINE from Pattern.compile method and you will see that result is

group=e

Because caret ^ was treated as start of entire input.

To know more about regex you may find useful my tutorial.

[Sudhanshu Mishra]

Thanks all for replies.
I was actually working on a problem statement which is "hide the first two characters of a string with length 4"
Input: abcd
Output : **cd

I came up with a regex
(?=^[\w]{4}$)[\w](?![\w]{0,2}$)

This basically checks whether the length is 4 first, then it checks if a letter is followed by a string with length greater than 2.
But what it is actually doing is *bcd.
I guess the next time it starts checking the regex it fails because now it is not on the start of the string.
But I am not sure what could be a workaround.
Any help is much appreciated.

Thanks in advance

[Rob Spoor]

Why not simply use String.length and String.substring?

[Sudhanshu Mishra]

design constraints
Have to do it by regex only

[Rob Spoor]

That's a stupid constraint. Using two simple methods from String is easier to write, easier to understand and probably more efficient as well.

[Winston Gutkowski]

Rob Spoor wrote:That's a stupid constraint.

Agreed. Unless it's for a CS class or exam.

@Sudhanshu: regexes are great, but they're NOT good for everything; and they're generally SLOWER than traditional logic.

As far as your original one is concerned, if you simply want the first character, then "^." will do the trick. And if you want to find the first non-space character, then "^[\s]*." (with appropriate brackets if you want to capture it).

But, I'm in complete agreement with Rob: Do it the way that's easiest for other people to understand. Regexes are arcane enough without adding a bunch of unnecessary logic to them.

Winston

[Bin Smith]

And if you want to find the first non-space character, then "^[\s]*."

For non-space characters we use \S but for space characters - \s.

[Winston Gutkowski]

Volodymyr Levytskyi wrote:For non-space characters we use \S but for space characters - \s.

Yes, but if you want to find the first non-space character, you first have to eliminate all the leading space ones.

Winston