Java String regex replace methods remove backslashes from replacement

Question

I just lost 1 hour to track what I consider not normal behaviour with replaceAll/replaceFirst in the String class.

If there is a backslash in the replacement String then they are removed when replacing. I then read that you can use Matcher.quoteReplacement(String) to create a proper replacement string,but my question is why? I can expect that the first argument should be escaped with Patter.quote(String) if you don't want the special meaning but I don't see a reason to change the replacement :(

Yes I will start using replace(CharSequence,CharSequence), just want to know why :)

Here is an example that clearly shows the "strange" behaviour:

public static void main(String[] args) {
    String out = "\\\\test\\\\";
    System.out.println(out);
    String result = "a".replaceAll("a", out);
    System.out.println(result);
}

note how second line is with only single backslashes as opposed to two like in the first line

docs.oracle.com/javase/7/docs/api/java/util/regex/… docs.oracle.com/javase/7/docs/api/java/util/regex/… — SLaks
– SLaks, Commented Oct 23, 2013 at 13:51
Yes I read the docs after seeing the problem and found about Matcher.quoteReplacement but this doesn't explain the fact that this seems to be against en.wikipedia.org/wiki/Principle_of_least_astonishment at least for me :( — nikolavp
– nikolavp, Commented Oct 23, 2013 at 13:56

anubhava · Accepted Answer · 2013-10-23 13:48:51Z

2

Yes it is true that backslash needs to be doubly escaped as the first argument in String#replaceAll.

Reason:

It is due to the fact that your replacement String can contain back-references like $1, $2 etc even replacement text is also processed by underlying regex engine hence the need for double escaping same as the first argument, as you have also found out.

answered Oct 23, 2013 at 13:48

anubhava

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SLaks · Accepted Answer · 2013-10-23 14:01:11Z

2

As has been mentioned, the replacement string uses $ characters to reference capture groups.

Therefore, to insert a literal $, you need to escape it with a \.

You can see this in the source.

I'm not sure why they didn't use $$ like .Net does.

answered Oct 23, 2013 at 14:01

SLaks

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