in reply to Array of ($1, $2, ...) in replacement part of s///?

Nope, @-, @+, and substr is the way. People periodically suggest a @& variable to do what you want, but that would invoke all the drawbacks of $&.

One handy feature is that when you say $s =~ s/re/expr/ge within expr, $s will be the original $s and the offsets in @- and @+ will be offsets into the original $s, even if earlier matches changed the length of $s.

Would be nice if $& (and the vaporware @&) were usable in the substitution part of s/// without penalty...

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Re: Re: Array of ($1, $2, ...) in replacement part of s///?
by flounder99 (Friar) on Jan 28, 2004 at 21:02 UTC
    Wow, I learn something new about perl every day.

    from perlvar: 

        After a match against some variable $var:

$` is the same as substr($var, 0, $-[0])
$& is the same as substr($var, $-[0], $+[0] - $-[0])
$' is the same as substr($var, $+[0])
$1 is the same as substr($var, $-[1], $+[1] - $-[1])
$2 is the same as substr($var, $-[2], $+[2] - $-[2])
$3 is the same as substr $var, $-[3], $+[3] - $-[3])

    [download]
    so I wrote a test case: 
    use strict;
for    my $regex (qr/(.)oo(.)ar(.)az/, qr/fo(.)barba(.)/) {
    my $str = "foobarbaz";
    $str =~ s/$regex/
        my @a;
        for my $index (1 .. $#+) {
            push @a, substr($str, $-[$index], $+[$index] - $-[$index])
+;
        }
        join "," , @a;
    /e;
    print "$str\n";
}

    [download]
    outputs 
    f,b,b
o,z

    [download]
    I'll have to remmember that one.

    --

    flounder

[reply]
[d/l]
[select]

      And now in the usual idiom as well.

      join
    "," ,
    map
        substr( $str, $-[$_], $+[$_] - $-[$_] ),
        1 .. $#+;

      [download]
[reply]
[d/l]
        Reminds me of the old BBS days. 
        join ",",map substr($str,$-[$_],$+[$_]-$-[$_]),1..$#+;#NO CARRIER

        [download]

        --

        flounder

[reply]
[d/l]

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