Ha! Got another solution. Using what chromatic previously suggested. I discovered that it is working, but it is ignoring the space in the '$1 ', so the string is not modified. However, if the replacement string is specified as '$1." "', because the /e modifier evalutes it as a Perl expression, it correctly puts the space after the value of $1.

So here's another version that works: 

#!/usr/bin/perl

 my $string="realllylongstringthatrefusestoend <A HREF=\"http://perlmo
+nks.org/images/blah/blah/blah\">\n";
 print splitter($string,"<.*?>",'(\S{18})','$1." "');
 sub splitter{
         my($string,$spliton,$find,$replace)=@_;
         my @array=split(/$spliton/,$string);
         my $i=0;
         my @splitters;
         my $str;
         while($string=~/($spliton)/g){
                 push @splitters,$1;
         }
         for (@array){
                s/$find/$replace/eeg; ;
                $str.=$array[$i];
                $str.=$splitters[$i];
                $i++;
         }
         $str;
 }

[download]
I believe this has the same security problems as evaluating with double quotes, because it allows the execution of arbitrary perl code.

--ZZamboni

In reply to RE: Handling scalars as regexs within a substitution. (Take 2) by ZZamboni
in thread Handling scalars as regexs within a substitution. (Take 2) by vroom

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