Anonymous Monk has asked for the wisdom of the Perl Monks concerning the following question:

Hi Monks, I want to do a nested replace. But I do not know how to code it.

I want to delete from string(= complete file with multiple lines) everything between "/*" and */, but all "\n" should be untouched.

My Idea was to search for "/*..*/ and than replace in a second step all characters not equal to \n with ""

Here my construct, which does not work :-( 
s:/\*([\0-\255]*?)\*/:($1=~s/[^\n]//g):g;
                      ^^^^^^^^^^^^^^^^ nested replace is seen as text 
+and not executed

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Can you tell me what I how to code it? By the way: Is there a shorter expression for : [\0-\255]? Thanks !!

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Re: nested replace
by choroba (Cardinal) on Mar 18, 2016 at 08:56 UTC
    If you want the dot to match anything including newlines, use the /s modifier. To evaluate the replacement as code, use the /e modifier. 
    use Syntax::Construct qr{ /r };
$string =~ s:/ \* (.*?) \*/ : $1 =~ s/[^\n]//gr :gsex;

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    In older Perls without /r : 

    $string =~ s:/ \* (.*?) \*/ : (my $x = $1) =~ s/[^\n]//g; $x :gsex;

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    ($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
}map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,

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      Thanks choroba, for your comments of the modifier. I used: 

      $string =~ s:/ \* (.*?) \*/ : (my $x = $1) =~ s/[^\n]//g; $x :gsex;

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      but it replaces the text within the "/*...*/" with the a position of the text. 
      ">/*x*/<"   is converted in ">1<" instead of "><"
">/*x\nx\nx*/<"  is converted in ">3<" instead of ">\n\n<"

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        Weird, works for me in 5.8.3, 5.10.1, 5.20.1, and 5.22.1. 
        $ perl -lwe ' $string = q<a/*z*/b>;
              $string =~ s: /\* (.*?) \*/ : (my $x = $1) =~ s/[^\n]//g
+; $x :gsex;
              print $string;
            '
ab

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        ($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
}map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,

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        You probably forgot the final "; $x" in the replacement part of the outer regex. Without it, the inner substitution returns the number of times it matched.

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        Sorry !

        Your code works correct ! Thanks!!

        In my code I did not have the last $x

        $string =~ s:/ \* (.*?) \*/ : (my $x = $1) =~ s/[^\n]//g; $x :gsex;
                                                          ^^

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Re: nested replace
by AnomalousMonk (Archbishop) on Mar 18, 2016 at 13:22 UTC

    Just a parenthetic note: From Perl version 5.14 on, the  /r modifier for  s/// and  tr/// operators allows them to return the substituted/transformed string for a slightly simpler expression: 

    c:\@Work\Perl\monks>perl -wMstrict -le
"my $c = qq{foo /* bar \n baz \n boff \n */ fum};
 print qq{<<$c>>};
 ;;
 $c =~ s{ /[*] (.*) [*]/ }{ $1 =~ s{[^\n]}{}xmsrg }xmsge;
 print qq{[[$c]]};
"
<<foo /* bar
 baz
 boff
 */ fum>>
[[foo


 fum]]

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    Works with  $1 =~ tr{\n}{}cdr as well. See  s/// in Regexp Quote-Like Operators, and  tr/// in Quote-Like Operators.

    Give a man a fish:  <%-{-{-{-<

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Re: nested replace
by Anonymous Monk on Mar 18, 2016 at 09:28 UTC

    Another alternative:

    use Regexp::Common 'comment';
$str=~s/($RE{comment}{C})/($a=$1)=~s#[^\n]##g;$a/ge;

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Re: nested replace
by Discipulus (Canon) on Mar 18, 2016 at 08:57 UTC
    i'm not an expert but i think you need an e as regex modifier.

    L*

    There are no rules, there are no thumbs..
    Reinvent the wheel, then learn The Wheel; may be one day you reinvent one of THE WHEELS.
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Re: nested replace
by Anonymous Monk on Mar 18, 2016 at 08:55 UTC

    generic "pattern" hahyaha 

     $var =~ s{(yada)}{OtherSubstitute("$1")}ge; 
sub OtherSubstitute {
  my( $in ) = @_;
##  $in =~ s{yada}{yada}g;
  my $numlines = $in =~ tr/\n/\n/; ## count
  return "\n" x $numlines;
}

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