Re: Recursion and ${*digit*} variables

On Fri, 14 Aug 2009 16:17:32 -0700 (PDT), "C.DeRykus" <derykus@xxxxxxxxx> wrote:

On Aug 14, 2:46 pm, "C.DeRykus" <dery...@xxxxxxxxx> wrote:
On Aug 14, 11:49 am, Mark <goo...@xxxxxxxxxxxxxxxx> wrote:



It appears that new ${*digit*} variables are not created when the same
subroutine is re-entered.  Is this to be expected?

use strict ;
use warnings;

my $line = 'abc';
doit();
exit;

sub doit {
#   local($1);  # adding this doesn't appear to have an effect
    $line =~ /(.)/ || die "no match";
    print "bef \$1  =$1\n";

    $line = substr($line,1);
    doit() if length($line);

    my $thing = $1; # WRONG! $1 has been changed by recursive call to
doit()

    print "aft \$1  =$1\n";

}

Output:
bef $1  =a
bef $1  =b
bef $1  =c
aft $1  =c
aft $1  =c
aft $1  =c

Expected output:
bef $1  =a
bef $1  =b
bef $1  =c
aft $1  =c
aft $1  =b
aft $1  =a

No, this came up in a recent thread. Even if a match fails
within a given scope, a previously successful backref isn't
cleared:

my $s='a';
for (1..2){
  print $s=~/(.)/ ? "match:$1" : "no match:$1";
  $s=substr($s,1);}

__END__

match:a
no match:a


Just to expand that explanation a bit more:

sub doit {
$line =~ /(.)/ || die "no match";

print "bef \$1 =$1\n"; # matches succeed initially
$line = substr($line,1); # as $line gets emptied
doit() if length($line); # and doit() recurses
my $thing = $1; # WRONG! ..

print "aft \$1 =$1\n";
}

By the time the recursion starts to unwind, $line has
been drained and the match fails. Because the final "c"
matched successfully though, $1 doesn't get cleared.
So you see the trailing series of "c" printed.

Matching had nothing to do with why his $1 contained 'c'
when the stack unwound.

The reason is he keeps reenterring the same scope block.
In terms of N vars, there is only one scope block so
the N vars keep thier last state.

N var state is not stacked, it only has scope.

-sln

.

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