Question

Possible Duplicate:
String.replaceAll() anomaly with greedy quantifiers in regex

I was writing code that uses Matcher#replaceAll and found following result highly confusing:

Pattern.compile("(.*)").matcher("sample").replaceAll("$1abc");

Now, I would expect the output to be sampleabc but Java throws at me sampleabcabc.

Does anybody have any ideas why?

Now, sure, when I anchor the pattern (^(.*)$) the issue goes away. Still I don’t know why the hell would replaceAll do a double replacement like that.

And to add insult to injury, following code:

Pattern.compile("(.*)").matcher("sample").replaceFirst("$1abc")

works as expected, returning just sampleabc.

Answer 1

It looks like it’s matching the empty string at the end of the input, for some reason. (I can see why it would match; I’m intrigued that it matches once and only once.)

If you change replaceAll("$1abc") to replaceAll("'$1'abc") the result is 'sample'abc''abc.

Note that if you change (.*) to (.+) then it works correctly, because it has to match at least one character.

The diagnosis is confirmed by this code:

Matcher matcher = Pattern.compile("(.*)").matcher("sample");

while (matcher.find()) {
    System.out.printf("%d to %d\r\n", 
                      matcher.start(), 
                      matcher.end());
}

… which outputs:

0 to 6
6 to 6