What does the "[^][]" regex mean?

Question

I found it in the following regex:

\[(?:[^][]|(?R))*\] 

It matches square brackets (with their content) together with nested square brackets.

Answer 1

[^][] is a character class that means all characters except [ and ].

You can avoid escaping [ and ] special characters since it is not ambiguous for the PCRE, the regex engine used in preg_ functions.

Since [^] is incorrect in PCRE, the only way for the regex to parse is that ] is inside the character class which will be closed later. The same with the [ that follows. It can not reopen a character class (except a POSIX character class [:alnum:]) inside a character class. Then the last ] is clear; it is the end of the character class. However, a [ outside a character class must be escaped since it is parsed as the beginning of a character class.

In the same way, you can write []] or [[] or [^[] without escaping the [ or ] in the character class.

Note: since PHP 7.3, you can use the inline xx modifier that allows blank characters to be ignored even inside character classes. This way you can write these classes in a less ambigous way like that: (?xx) [^ ][ ] [ ] ] [ [ ] [^ [ ].

You can use this syntax with several regex flavour: PCRE (PHP, R), Perl, Python, Java, .NET, GO, awk, Tcl (if you delimit your pattern with curly brackets, thanks Donal Fellows), ...

But not with: Ruby, JavaScript (except for IE < 9), ...

As m.buettner noted, [^]] is not ambiguous because ] is the first character, [^a]] is seen as all that is not a a followed by a ]. To have a and ], you must write: [^a\]] or [^]a]

In particular case of JavaScript, the specification allow [] as a regex token that never matches (in other words, [] will always fail) and [^] as a regex that matches any character. Then [^]] is seen as any character followed by a ]. The actual implementation varies, but modern browser generally sticks to the definition in the specification.

Pattern details:

\[          # literal [ (?:         # open a non capturing group     [^][]   # a character that is not a ] or a [   |         # OR     (?R)    # the whole pattern (here is the recursion) )*          # repeat zero or more time \]          # a literal ] 

In your pattern example, you don't need to escape the last ]

But you can do the same with this pattern a little bit optimized, and more useful cause reusable as subpattern (with the (?-1)): (\[(?:[^][]+|(?-1))*+])

(                     # open the capturing group     \[                # a literal [         (?:           # open a non-capturing group             [^][]+    # all characters but ] or [ one or more time           |           # OR             (?-1)     # the last opened capturing group (recursion)                       # (the capture group where you are)         )*+           # repeat the group zero or more time (possessive)     ]                 # literal ] (no need to escape) )                     # close the capturing group 

or better: (\[[^][]*(?:(?-1)[^][]*)*+]) that avoids the cost of an alternation.