Why does the look-behind expression in this regex not have an “obvious maximum length”?

Question

Given a string containing some number of square brackets and other characters, I want to find all closing square brackets preceded by an opening square bracket and some number o

Answer 1

\[ is only a single character, so it seems like the obvious maximum length should be 1 + whatever the obvious maximum length was of the look-behind group in the first expression. What gives?

That's the point, "whatever the obvious maximum length was of the look-behind group in the first expression", is not obvious. A rule of fist is that you can't use + or * inside a look-behind. This is not only so for Java's regex engine, but for many more PCRE-flavored engines (even Perl's (v5.10) engine!).

You can do this with look-aheads however:

Pattern p = Pattern.compile("(?=(\\[[a-z]+]))");
Matcher m = p.matcher("] [abc] [123] abc]");
while(m.find()) {
  System.out.println("Found a ']' before index: " + m.end(1));
}

(I.e. a capture group inside a look ahead (!) which can be used to get the end(...) of the group)

will print:

Found a ']' before index: 7

EDIT

And if you're interested in replacing such ]'s, you could do something like this:

String s = "] [abc] [123] abc] [foo] bar]";
System.out.println(s);
System.out.println(s.replaceAll("(\\[[a-z]+)]", "$1_"));

which will print:

] [abc] [123] abc] [foo] bar]
] [abc_ [123] abc] [foo_ bar]

Answer 2

 "^[^\[]*\[[^\]]*?(\])"

is the group(1) what you want?