Once again, you have correctly corrected me. :-) Thanks.
I didn't happen to notice that OP was passing a reference, so, as you say, the issue was moot. On the other hand, whilst you are right that Perl passes by reference and I am grateful for the correction, the usual semantics of parameter
usage does involve copies that can clobber a passed in tied scalar:
In this code sample, which compares the usage of a parameter array element directly ($_[0]) with one common usage of the array ($param = shift @_)
use strict;
use warnings;
{
package TiedObject;
sub TIESCALAR {
my $sClass = shift;
return bless([ @_ ], $sClass);
}
sub STORE {}
sub FETCH { shift->[0]; }
}
my $sErr;
tie $sErr, 'TiedObject', 5, "More data";
printTiedObject($sErr);
sub printTiedObject {
my $oErr = tied($_[0]);
#does not copy scalar, prints out "...is tied to..."
print "using \@_ directly\n";
if (defined($oErr)) {
print "\$_[0] is tied to <$oErr>\n";
} else {
print "\$_[0] is not tied";
}
#copies scalar, prints out "..is not tied"
print "\nafter \$param0 = shift \@_\n";
my $param0 = shift @_;
$oErr = tied ($param0);
if (defined($oErr)) {
print "\$param0 is tied to <$oErr>\n";
} else {
print "\$param0 is not tied\n";
}
}
we get the following output
using @_ directly
$_[0] is tied to <TiedObject=ARRAY(0x814ed48)>
after $param0 = shift @_
$param0 is not tied
Best, beth
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