PERL Beginners - Multidimensional array / Passing an array to a sub

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Multidimensional array / Passing an array to a sub

Martin Tournoij

2006-07-30, 3:57 am

I'm currently massivly

by multidimensional arrays and passing =

arrays to subs.

Why is it that @_[1] isn't properly copied to @a?

I'm not sure what exactly I'm doing wrong here, probably something simpl=
e, =

but I can't find anything that works in the manpages or on the site...

Music($var1, \@array);

sub Music
{
print $_[1][1]; # works
my @a =3D @_[1] # Gives warning, should be written as $_[1]

print $a[1]; # Outputs ARRAY(#memaddr)
print $a[1][1]; # Works

for (@a)
{
print; # Outputs ARRAY(#memaddr)
}

for (@_[1])
{
print; # Also outputs ARRAY(#memaddr)
}
}

Charles K. Clarkson

2006-07-30, 3:57 am

Martin Tournoij wrote:

: I'm currently massivly

by multidimensional arrays and
: passing arrays to subs.
:
: Why is it that @_[1] isn't properly copied to @a?

It is properly copied. Because @_[1] one is a scalar and
@a is an array the reference to the passed array in stored in
$a[0].

: Music($var1, \@array);
:
: sub Music
: {
: print $_[1][1]; # works
: my @a = @_[1] # Gives warning, should be written as $_[1]

So stop writing it as @_[1], silly. :)

my $var = shift;
my $array_ref = shift;

: print $a[1]; # Outputs ARRAY(#memaddr)

print @{ $array_ref };

: print $a[1][1]; # Works
:
: for (@a)

for ( @{ $array_ref } ) {

HTH,

Charles K. Clarkson
--
Mobile Homes Specialist
Free Market Advocate
Web Programmer

254 968-8328

Don't tread on my bandwidth. Trim your posts.

Thomas J.

2006-07-30, 3:57 am

Martin Tournoij schrieb:

>
> Music($var1, \@array);
>
> sub Music
> {
> print $_[1][1]; # works
> my @a = @_[1] # Gives warning, should be written as $_[1]

yes, it should be written as $_[1] , if you want the ref of your
@array.
perl doesnt know what you want here, but i think you want:

my @a = @{ $_[1] };

or something similar.
all of your errors will gone after that.

>
> print $a[1]; # Outputs ARRAY(#memaddr)

with my suggestion the line one down wont work anymore
the line above will probably do what you want.

> print $a[1][1]; # Works
>
> for (@a)
> {
> print; # Outputs ARRAY(#memaddr)
> }
>

with my suggestion the line one down has also be changed to:
for ( @{ $_[1] } )

> for (@_[1])
> {
> print; # Also outputs ARRAY(#memaddr)
> }
> }

if you want the structure of a reference you have to dereference it.
see: perldoc perlref

hope this helps

Thomas

John W. Krahn

2006-07-30, 7:56 am

Martin Tournoij wrote:
> I'm currently massivly

by multidimensional arrays and passing
> arrays to subs.
>
> Why is it that @_[1] isn't properly copied to @a?
>
> I'm not sure what exactly I'm doing wrong here, probably something
> simple, but I can't find anything that works in the manpages or on the
> site...
>
> Music($var1, \@array);
>
> sub Music
> {
> print $_[1][1]; # works

$_[1] contains a reference to @array so $_[1][1] is the same as $array[1] (or
the second element of @array.)

> my @a = @_[1] # Gives warning, should be written as $_[1]

@_[1] is an array slice, the warning is because the list [1] has only one
element, it's the same as saying "$a[0] = @_[1]". If you want to copy @array
to @a then you have to dereference it properly:

my @a = @{ $_[1] }

But if you are going to copy the array anyway you could just do it like this:

Music($var1, @array);

sub Music
{
my ( $var, @a ) = @_;

John
--
use Perl;
program
fulfillment

Paul Lalli

2006-07-30, 6:57 pm

Charles K. Clarkson wrote:
> Martin Tournoij wrote:
>
> : I'm currently massivly

by multidimensional arrays and
> : passing arrays to subs.
> :
> : Why is it that @_[1] isn't properly copied to @a?
>
> It is properly copied. Because @_[1] one is a scalar

No. It's not. And that's why you get the warning.

@_ is an array.
$_[1] is a scalar - the second element of the array @_
@_[1] is an array slice - the list containing the second element of the
array @_

Paul Lalli

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