Stringification of a compiled regex produces something that can be used in other qr//, m// or s/// without changing the meaning of the pattern. Adding (?^:...) or similar is necessary to achieve that.

my $re = qr/a/;

say "a" =~ /$re/    || 0;   # 1
say "A" =~ /$re/    || 0;   # 0
say "a" =~ /$re/i   || 0;   # 1
say "A" =~ /$re/i   || 0;   # 0

my $pat1 = "(?^:a)";

say "a" =~ /$pat1/  || 0;   # 1
say "A" =~ /$pat1/  || 0;   # 0
say "a" =~ /$pat1/i || 0;   # 1
say "A" =~ /$pat1/i || 0;   # 0

my $pat2 = "a";

say "a" =~ /$pat2/  || 0;   # 1
say "A" =~ /$pat2/  || 0;   # 0
say "a" =~ /$pat2/i || 0;   # 1
say "A" =~ /$pat2/i || 0;   # 1  XXX
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That said, you can get the unmodified pattern (and the flags) using re's re::regexp_pattern.

my ( $pat, $flags ) = re::regexp_pattern( $re );
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You can rebuild the original compiled regex using the following:

my $re = eval "no re; qr/\$pat/$flags";
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It adds the ?^: at the start. What is the proper way to get rid of it?

Try Data::Dump pp to serialize regex or something from re or B to get original qr

maybe use https://perldoc.perl.org/re#regexp_pattern($ref)

my ($pat, $mods) = regexp_pattern($ref);
my $parentless = ( $pat =~ m/[^:]+:(.*?)\)$/s )[0];
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Thank you very much!
One question - why the second line is needed? It looks like $pat does what I need.

My cell phone doesnt have perl, the doc example implied the (^:) being returned

You used a lot of escaping of \/ in your regex:

> "regex" => qr/^\/a\/.*?\/b\/.*\/pkgs\/([^\/]*)?\/([^\/]*)?\/.*/

Are you aware that qr and his siblings of "quote-like-operators" allow other delimiters?

Like qr(...) or qr~...~ and so on?