multiple assignments with a HoHoA

[Abhinav]

Hi,

I have a hash variable, %table, which I populate as

my %table;
$table{'A'}{'B'}=['C','D','E'];

I now do

my $val1;
my $val2;
my $val2;

This :

($val1, $val2, $val3) = $table{'A'}{'B'}; #causes $val1, $val2 to
be undef, #$val3 = ARRAYREF

However, this :

$val1 = $table{'A'}{'B'}[0];
$val2 = $table{'A'}{'B'}[1];
$val3 = $table{'A'}{'B'}[2]; # Does the Right Thing.

Could someone point out what silly mistake I am committing in the 1st case,
and how to rectify it ?


Thanks

Abhinav

 

[Gunnar Hjalmarsson]

I have a hash variable, %table, which I populate as

my %table;
$table{'A'}{'B'}=['C','D','E'];

I now do

my $val1;
my $val2;
my $val2;

This :

($val1, $val2, $val3) = $table{'A'}{'B'}; #causes $val1, $val2 to
be undef, #$val3 = ARRAYREF

Suppose you mean that $val1 is assigned the array reference, not $val3.

Could someone point out what silly mistake I am committing in the 1st
case,

You don't dereference your array reference.

and how to rectify it ?

Dereference your array reference.

perldoc perlreftut

 

[Mark Clements]

I have a hash variable, %table, which I populate as

my %table;
$table{'A'}{'B'}=['C','D','E'];

OK: this is actually syntactic sugar for

$table{A}->{B} = [ 'C', 'D', 'E'];
# note you'd probably use qw here instead ie [qw( C D E )]

this sets $table{A}->{B} to be an arrayref

Perl allows you to specify multiple dereferences without the ->,
supposedly to make it easier to read

ie

$table{A}{B}{C} is the same as $table{A}->{B}->{C} but not the same as
$table->{A}->{B}->{C}.

($val1, $val2, $val3) = $table{'A'}{'B'}; #causes $val1, $val2 to
be undef, #$val3 = ARRAYREF

it should set $val1 *not* $val3: $table{A}->{B} is a scalar value ie an
arrayref, so the right-hand side only returns one value, which is
assigned to the first item in the list on the lhs.

However, this :

$val1 = $table{'A'}{'B'}[0];
$val2 = $table{'A'}{'B'}[1];
$val3 = $table{'A'}{'B'}[2]; # Does the Right Thing.

syntactic sugar is coming into play again.

$val1 = $table->{A}->{B}->[0];

and so on.

Could someone point out what silly mistake I am committing in the 1st
case, and how to rectify it ?

Not silly so much as missing some of the subtleties of perl data
structures - you need to check out

man perldsc

Mark

 

[Tore Aursand]

my %table;
$table{'A'}{'B'}=['C','D','E'];

I now do

my $val1;
my $val2;
my $val2;

I guess you mean '$val3' on that last one. No need to declare them before
you actually use them;

my ( $val1, $val2, $val3 ) = ...;

This :

($val1, $val2, $val3) = $table{'A'}{'B'}; #causes $val1, $val2 to
be undef, #$val3 = ARRAYREF

You need to dereference your array reference;

my ( $val1, $val2, $val3 ) = @{$table{'A'}{'B'}};

 

[Abhinav]

my %table;
$table{'A'}{'B'}=['C','D','E'];

I now do

my $val1;
my $val2;
my $val2;

I guess you mean '$val3' on that last one. No need to declare them before
you actually use them;

Oops ..yes ..

my ( $val1, $val2, $val3 ) = ...;




You need to dereference your array reference;

my ( $val1, $val2, $val3 ) = @{$table{'A'}{'B'}};

Thanks ! I skimmed through perldsc before my OP, but I did not grasp it. It
is much clearer now.

Even the above didn't work initially, thanks to precedence issues as I was
using the || operator directly after this assignment.

Thanks to everyone for the useful replies

Regards
Abhinav

P.S : The reason for this post was this week's QOTW. While it sounded a bit
complex, it was actually easy, and newbies might want to go through it. It
is available at http://perl.plover.com/qotw/r/current.txt (This is quiz #24)

 

[Shawn Corey]

Hi,

($val1, $val2, $val3) = @{ $table{'A'}{'B'} };

--- Shawn

 

[Gunnar Hjalmarsson]

I skimmed through perldsc before my OP, but I did not grasp it.

Of course not. The docs about references and data structures is
nothing you just "skim"!