in reply to How to start regexp from a X position

Here a way of doing it with pos: 
my $x = "<0>";
my $pos = 0;
for (qw {<1> <2> <3>}) {
    pos($x) = $pos;
    $x =~ s/\G(<\d+>)/$1$_/;
    $pos += length $1;
}
say $x;
__END__
<0><1><2><3>

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Re^2: How to start regexp from a X position
by pkt (Initiate) on Oct 07, 2008 at 12:20 UTC

    Thank!! great!!!, you open my mind, really i was knew that the solution is around that but i couldn't understand the documentation, perfect!!!!.

    One last question, how works the modifier c and \G in a regexp

    Could i do something like this:

    my $x = "<0>";

for (qw {<1> <2> <3>}) {
    if ($x =~ /<\d+>/gc) {
        $x =~ s/\G(<\d+>)/$1$_/;
    }
}

say $x;

    [download]

    Really thank JavaFan

[reply]
[d/l]

      Thank!! great!!!, you open my mind,

      You picked the worst of the presented solutions solely because it used pos as you initially believed it should. That's the very definition of a closed mind.

[reply]
[d/l]

        Don't angry with me!!!, yes you right i'm a little closed mind, and you and JavaFan help me open it, when you said use .* i said me couldn't be more stupid it's right!!!, i really fight before posting here for many hours and i cuoldn't find a solution, becouse of that i reasearch about not common at least for me functio, variable and operators like pos, $&, \G, etc, etc, but finally you right :-)

        Thank to both

[reply]
      I've no idea how to incorporate the /c modifier. It'll be pointless for two reasons. First, one sets pos() before matching, so who cares about the effect of /c. Second, /c would only kick in if you add a substring foo to $x, and the next time around, foo isn't there.
[reply]

        Ok, JavaFan

[reply]

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