Is it possible to do pass by reference in Perl?

[Anonymous Monk]

Is it possible to do pass by reference in Perl?

Originally posted as a Categorized Question.

[chromatic]

Prepending a variable with a backslash (\) creates a reference:

my $scalar = "this is a scalar";
my @array = qw( this is my array );
my %hash = (
    'this' => 'my',
    'hash' => 'mine',);

print_all(\$scalar, \@array, \%hash);

sub print_all {
    my $scalar_ref = shift;
    my $arr_ref = shift;
    my $hash_ref = shift;

    print "Scalar: ", $$scalar_ref, "\n";
    print "Array: ", join ' ', @$arr_ref, "\n";
    print "Hash: ", each %$hash_ref, "\n";
}

See perlman:perlref for the juicy details.

[Russ]

One might argue that Perl always does Pass-By-Reference, but protects us from ourselves.

@_ holds the arguments passed to a subroutine, and it is common idiom to see something like:

sub mySub{
  my $Arg = shift;
}

sub mySub2{
  my ($Arg) = @_;
}

Why is that? Why don't we just use the @_ array directly?

First, there is the laudable goal of more readable code, which is sufficient reason, in itself, to rename variables away from cryptic things like $_[3]. But really, we copy values out of @_ because (from the man page) "its elements are aliases for the actual scalar parameters."

In short, this means that if you modify an element of @_ in your subroutine, you are also modifying the original argument. This is almost never the expected behavior! Further, if the argument is not updatable (like a literal value, or a constant), your program will die with an error like "Modification of a read-only value attempted."

Consider:

sub test{
  $_[0] = 'New Value';
}

my $Var = 'Hi there';
print "$Var\n";
test ($Var);
print "$Var\n";

will print out:

Hi there
New Value

So, yes, you can do pass-by-reference in Perl, even without backslashes; but it is almost always better (some would leave out the "almost" in this statement) to make your caller explicitly pass you a reference if you intend to modify a value.

[btrott]

... And, as would follow from chromatic's answer, if you modify the references within a subroutine, you'll also modify the values those references point to, outside the subroutine.

my $str = "foo";

print $str, "\n";
change(\$str);
print $str, "\n";

sub change {
    my $ref = shift;
    $$ref = "bar";
}

[tilly]

In addition to the answers above it is possible to pass arrays and hashes by reference without using a \. The way to do this is to define a function with a prototype as explained in perlsub.

For a more complete explanation I can recommend FMTYEWTK About Prototypes, which says in detail what they are, where they are buggy, and the various design flaws that make them something to avoid on the whole.

[bradcathey]

In fact, passing by reference is a must to preserve the values in separate data structures, even an array and scalar. For instance:

my @array = qw ( 1 2 3 4 5 );
my $num = 6;

&testsub ( @array, $num );

sub testsub {
   my ( @list, $single ) = @_;
   print @list;
}

will print:

123456

Passing those values via the @_ flattens all values into one long list. However:

my @array = qw ( 1 2 3 4 5 );
my $num = 6;

&testsub ( \@array, $num );

sub testsub {
   my ( $list, $single ) = @_;
   print @$list;
}

will print:

12345

where passing the array by ref will keep the values from all merging into the single list.

[ariels]

*Only* scalars can be passed to a sub as parameters, and they can *only* be passed by reference. If you modify @_, the actual variable passed in gets modified! @_ is really a list of aliases to the argument list.

Try it:

sub foo { $_[0] = 17 }
my $x=0;
foo $x;
print "$x\n"

And foo @x does what you'd expect given the aliasing.

Of course, the usual course of action is to copy them into my variables. Which is why it's perceived as call by value.

[stefp]

It is only possible to pass scalars by reference to a sub.

Parameters are passed by value in Perl. But we have an operator, the backslash, that returns reference to any variable (scalar or composite). So the net result is that we have a mechanism to pass parameter by reference.

Admittedly, this not very clean, because the formal parameter is declared as a scalar even if it is a reference to a non scalar. I guess that was the point that tried to convey the sentence: It is only possible to pass scalars by reference to a sub.

sub foo { my $a = shift;
  push @$a, "foobar";
}

my @a = ();
foo \@a;     # passing a value that is a reference to @a
print $a[1]; # prints "foobar"

tilly messages me: another example of how to pass arrays by reference is by using prototypes. Unlike using \ in user code, that is transparent.. So Perl has call by reference after all (even if it was quite a late addition?).

-- stefp

[RedDragon]

It is possible to pass by reference in perl. The argument list u get inside ur subroutine via @_ are implicit reference to values that were passed in from inside from outside.i.e. if u pass in a list of strings, inside the body of the sub,change those strings,then they will be modified outside the subroutine.

Originally posted as a Categorized Answer.

[RedDragon]

It is possible to pass by reference in perl. The argument list u get inside ur subroutine via @_ are implicit reference to values that were passed in from inside from outside.i.e. if u pass in a list of strings, inside the body of the sub,change those strings,then they will be modified outside the subroutine.

sub ted {
my($add,$sub,$mul) = @_;
...
..
}