Re^3: Challenge: Transforming markups

[LanX]

OK, but the problem here are nested capture groups.

you can't rely on $2 or $3 to be correct if $named is something like [[ ($http) ][ ($title) ]]

[Anonymous Monk]

If you just don't like referring to the global %-, maybe something like:

s/$regex/transform(%-)/ge
sub transform {
   my %matches = @_;
   ...
}

Or you can do a second match.

s/($regex)/transform($1)/ge
sub transform {
   local $_ = shift;
   if (/^(\[\[($http)]\[($title)]])/) { return transform_named($2,$3) 
+}
   elsif (/^($http)/) { return transform_http($1) }
   ...
}

Or you might use something like this.

while (1) {
   if (/\G($named)/gc) { print transform_named($1) }
   elsif (/\G($http)/gc) { print transform_http($1) }
   ...
   elsif (/\G(.)/gcs) { print $1 }
   else { last }
}

[LanX]

> Or you might use something like this.

while (1) {
   if (/\G($named)/gc) { print transform_named($1) }
   elsif (/\G($http)/gc) { print transform_http($1) }
   ...
   elsif (/\G(.)/gcs) { print $1 }
   else { last }
}

Thanks I ended up using this approach, it's the most practical in my case! ¹

Actually I knew and tried it before w/o luck, turned up that the /\G(.)/gcs was crucial.

I remember \G to be tricky (or even buggy) in edge cases,

(Don't really know why I need /s since there are no line breaks in single lines? ...have still to debug whats going on with pos here...)

But it works now, thanks! =)

¹) The code in the OP might be a good way to design a parser generator but I dont need the complexity and the extra speed of one big regex.

[hdb]

This can be easily fixed by using named backreferences. \k<named> would refer to a match of the pattern (?<named>...).