反向引用在 tcl 正则表达式中不起作用

Question

I am new to regular expressions and tcl and am facing very basic issue from a long time now.

I am given with the task to find all the characters in given word, whose immediate next character is not identical to this character. I have written following tcl snippet to achieve this:

set str "goooo";
set lst [regexp -all -inline {(\w)[^\1]} $str];
puts $lst

I am getting following error:

couldn't compile regular expression pattern: invalid escape \ sequence
    while executing
"regexp -all -inline {(\w)[^ \1]} $str"

Is there any other way to use backreferencing in tcl?

Answer 1

Backreferences cannot be used inside bracket expressions in any regex flavor. [^\1] matches any char but a \x01 char. This happens so because bracket expressions are meant to use exact literal characters or ranges of them .

In your case, you can remove all chunks of repeated chars with (\w)\1+ (while replacing with the same single char using the \1 backreference in the replacement pattern) and then extract the word chars:

set lst [regexp -all -inline {\w} [regsub -all {(\w)\1+} $str {\1}]];

See the online demo :

set str "sddgoooo";
set lst [regexp -all -inline {\w} [regsub -all {(\w)\1+} $str {\1}]];
puts $lst

Output:

s d g o

Note that in other regex flavors, you could use a regex with a negative lookahead: (\w)(?!\1) (see this regex demo ). The (?!\1) negative lookahead matches a location that is not immediately followed with Group 1 value. Unfortunately, Tcl regex flavor - although Tcl AREs generally support lookaheads - does not support lookaheads with backreference inside them.