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

given string containing number of square brackets , other characters, want find closing square brackets preceded opening square bracket , number of letters. instance, if string is

] [abc] [123] abc]

i want find second closing bracket.

the following regex

(?<=[a-z]+)\]

will find me second closing bracket, last one:

] [abc] [123] abc]

since want find first one, make obvious change regex...

(?<=\[[a-z]+)\]

...and "look-behind group not have obvious maximum length near index 11."

\[ single character, seems obvious maximum length should 1 + whatever obvious maximum length of look-behind group in first expression. gives?

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eta: it's not specific opening bracket.

(?<=a[b-z]+)\]

gives me same error. (well, @ index 12.)

\[ single character, seems obvious maximum length should 1 + whatever obvious maximum length of look-behind group in first expression. gives?

that's point, "whatever obvious maximum length of look-behind group in first expression", not obvious. rule of fist can't use + or * inside look-behind. not java's regex engine, many more pcre-flavored engines (even perl's (v5.10) engine!).

you can look-aheads however:

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

(i.e. capture group inside ahead (!) can used end(...) of group)

will print:

found ']' before index: 7

edit

and if you're interested in replacing such ]'s, this:

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

which print:

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