Jupiterhost.Net wrote:
>
> #!/usr/bin/perl
>
> use strict;
> use warnings;
>
> my $howdy = 'hello';
> push(@ref_to_destroy, \$howdy);
> for(@refs_to_destroy) {
> my $r = ref($_);
> if(defined $r) {
> close $_ if $r eq 'IO';
> undef $_ if $r eq 'SCALAR';
> }
> }
> print "I still have howdy -$howdy-\n" if defined $howdy;
>
>
> If I do undef ${ $_ } if ... then it works (IE you get no "I still
> have" message..
> Which means I have to do @{ $_ } %{ $_ } etc also... The etc part is the
> one throwing me now :)
>
i am little confuse, do you want to loose the ref $howdy is pointing to so
Perl can gc the ref or undef $howdy itself if $howdy is a ref? your method
dereference $howdy first and then undef whatever it's pointing to. you can
look at it like:
$howdy -> $ref -> $scalar
your undef ${$howdy} basically means:
$howdy -> $ref $scalar
which breaks the ref from $ref to $sclar but $howdy is untouched. for
example:
[panda]# perl -le '$_=\\$.; undef $$_; print "r" if ref $_'
r
[panda]#
notice $_ is still a ref but if you do:
[panda]# perl -le '$_=\\$.; undef $_; print "r" if ref $_'
[panda]#
nothing is print because you end up with:
$_ $ref -> $scalar
now if $ref and $scalar has no more reference, Perl can gc them. your
version ends up:
$_ -> $ref $scalar
and if $scalar has no more reference, Perl can gc it but Perl can't gc $ref
because $_ still points to it. if you don't want the undef-ness of my
version, you can always:
[panda]# perl -le '$_=\\$.; $_ = 0; print defined $_'
1
[panda]#
so that after you loose the ref, $_ is still defined.
finally, notice your undef $$_ is also bugous:
[panda]# perl -e '$_ = \\1; undef ${$_}'
Modification of a read-only value attempted at -e line 1.
[panda]#
you don't want that do you? :-)
at the end, it's your call and it depends on how you set up your ref and
what you want to do with it.
david
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