Re^2: nested replace

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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Re^3: nested replace by choroba (Cardinal) on Mar 18, 2016 at 10:09 UTC
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 $string; ' ab` [download] ($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, [download]	[reply] [d/l] [select]
Re^3: nested replace by Anonymous Monk on Mar 18, 2016 at 10:07 UTC
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.	[reply] [d/l]
Re^3: nested replace by Anonymous Monk on Mar 18, 2016 at 10:13 UTC
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; ^^` [download]	[reply] [d/l]