regex question

[Janki Shah]

Matcher m5 = Pattern.compile("\\\\d").matcher("\1d\2df\3");
		while(m5.find())
		{
			System.out.println(m5.start() + " " +m5.group());
		}

It does not print anything. I was expecting to get these groups: \1 \2 \3

[Matthew Brown]

Did you really mean "\1d\2df\3", or should that be "\\1d\\2df\\3"?

[Janki Shah]

Hi Matthew, it does not print anything in both the cases. and I am confuse in understanding the difference in both.
What I trying to find is the patter with backslash and followed by single digit.

[Jelle Klap]

Needs more backslashes
The backslash is treated as an escape character by both the regular expression engine and Java.
Starting with you pattern definition, if your intent is to match a backslash followed by a single digit, you need \\\\ (four backslashes) to match the single backslash.
Of course that leaves the single digit, which can be represented by the escape character \d BUT seen as Java will also treat that \ as the start of an escape charcater, you need to escape it by adding another \, yeielding \\d.
Putting that toghether gives you the pattern \\\\\\d (six backslashes in there).
Now, that will fix the pattern, but it still won't yield a result, because there's something amiss with the literal String value you're trying to match against.
Why don't you try figuring out why that is from here

Edit: ugh, too slow.

[Janki Shah]

import java.util.regex.Matcher;
import java.util.regex.Pattern;

public class Regex4 {

	public static void main(String[] args) {
	Matcher m5 = Pattern.compile(args[0]).matcher(args[1]);
		while(m5.find())
		{
			System.out.println(m5.start() + " " +m5.group());
		}
		
	}
}

 Java Regex4 "\\\d" "\\1sd\\3" 
1 \1
6 \3

If I do it with notepad++ and use the args[0] and args[1] at run time i have to pass these arguments args[0] = "\\\d" to find bachslsh and followed by a single digit. and If I use eclipse I use"\\\\d" so why this difference and also when use command line to pass pattern and matcher it works fine but not in eclipse.
Can you help in finding what am doing wrong? Thank you.

[Matthew Brown]

Remember that \ is a special character in a Java string literal. Therefore, every time you include it in a string literal you need to escape it with another slash.

Since it's also a special character in regular expressions, when matching against a real slash you need to escape it. So if you are trying to put such a regular expression in a string literal you have to add both levels of escaping, hence the slash-happy example Jelle gave.

Input from the command line doesn't have the same issue.

[Janki Shah]

Hi Jelle, Yes I found it and works fine Thank you.
Can you explain one more thing that why is it difference when I use notpad and then pass arguments at run time vs. eclipse?

Matcher m5 = Pattern.compile("\\\\\\d").matcher("\\\\1d\\\\2df\\\\3");

Jelle Klap wrote:Needs more backslashes
The backslash is treated as an escape character by both the regular expression engine and Java.
Starting with you pattern definition, if your intent is to match a backslash followed by a single digit, you need \\\\ (four backslashes) to match the single backslash.
Of course that leaves the single digit, which can be represented by the escape character \d BUT seen as Java will also treat that \ as the start of an escape charcater, you need to escape it by adding another \, yeielding \\d.
Putting that toghether gives you the pattern \\\\\\d (six backslashes in there).
Now, that will fix the pattern, but it still won't yield a result, because there's something amiss with the literal String value you're trying to match against.
Why don't you try figuring out why that is from here

Edit: ugh, too slow.

[Jelle Klap]

Matthew summed it up nicely in the reply above yours

[Janki Shah]

Thank you Jelle and Matthew. one last question.

Matcher m5 = Pattern.compile("\\\\\\d").matcher("\\\\1\\\\2\\\\3");
		while(m5.find())
		{
			System.out.println(m5.start() + " "  + m5.end() + " "+m5.group());
		}

output

1 3 \1
4 6 \2
7 9 \3

how the start index and end index work here?

[Matthew Brown]

Well, remember that because of the escaping, the actual string you're matching against is \\1\\2\\3 (you'd have got matches with pairs of slashes as well). So the following indicates the position of the matches

\\1\\2\\3
 ^- ^- ^-
0123456789

The start index is the start of the group (zero indexed), and the end index is the position immediately after the group.

[Janki Shah]

Thank you Matthew. Understood the point.