On 12/10/05, $Bill Luebkert <[EMAIL PROTECTED]> wrote:
> Chaddaï Fouché wrote:
>
> > $Bill Luebkert wrote:
> >
> >
> >>Perl normally passes by value .
> >>
> >
> > That is unaccurate : Perl always passes by reference (or rather by
> > alias), but we do the copy ourselves when we handle the arguments array :
> > @_ always has aliases on the arguments.
>
> The fact that Perl uses an aliasing method does not change the fact that
> it passes by value rather than by reference. You can modify the vrbl
> that an arg is aliased to, but that doesn't make it a reference. Many
> of us prefer not to create/utilize that condition though.
> The fact that you can modify the vrbl passed to a sub from in the sub without
> using a reference still doesn't mean that it's passed by reference.
No. You are mistaken. That is exactly what passing by reference means.
Example:
sub Increment($){ $_[0]++ };
If you want to protect your passed arguments from possible modification,
make a copy of them before submitting them:
my @args=($SomeVariable);
&Increment(@args); # $SomeVariable will not be touched
You could even write a byvalue wrapper if you want, something like
sub ByValue(\&;){
my $coderef = shift;
my @argvals = @_; # copying arg refs to new array, by value, or
at least COW
&$coderef(@argvals);
};
then you could call untrusted subroutines like so
@results = ByValue(\&Untrusted::module::baz, $one, $two);
without worring that $one and $two might get touched.
although this might (or might not) be clearer:
@results = &Untrusted::module::baz(my($notone,$nottwo)=( $one, $two));
--
David L Nicol
I'll see your time division multiplexing and
raise you an asynchronous trap
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