Perl: Assigning an array to a hash

Question

This syntax works:

$b{"x"} = [1,2,3];
pp %b;
# Displays ("x", [1, 2, 3])

But I need to be able to dynamically create the array's contents and assign it later. This does not work; help, what's the obvious part I'm missing?

@a = [1,2,3];
$b{"x"} = @a;
pp %b;
# Shows up as ("x", 1) ... not what I want or expected.

Tried these variations, too.

$b{"x"} = [@a];  # ("x", [[1, 2, 3]])  ...close

$b{"x"} = \@a;   # ("x", [[1, 2, 3]])  

$b{"x"} = [\@a]; # ("x", [[[1, 2, 3]]])  

$b{"x"} = %a;    # ("x", 0)

$b{"x"} = $a;    # ("x", undef)

$b{"x"} = [$a];  # ("x", [undef])

$b{"x"} = @{@a};  # ("x", 0)

And, ideally, I'd like to be able to get the array back out later as an array.

My "pretty print" function came from CPAN -- use Data::Dump qq/pp/; — Walt Stoneburner
– Walt Stoneburner, Commented Mar 21, 2011 at 23:10

cjm · Accepted Answer · 2011-03-21 23:15:53Z

45

The part you're missing is that @a = [1,2,3] doesn't make an array with 3 elements. It makes an array with one element which is an arrayref.

You meant @a = (1,2,3).

To assign that array to a hash element, you'd use either $b{"x"} = [@a] or $b{"x"} = \@a, depending on what you're trying to do. [@a] makes a new arrayref containing a copy of the current contents of @a. If the contents of @a change after that, it has no effect on $b{x}.

On the other hand, \@a gives you a reference to @a itself. If you then change the contents of @a, that will be visible in $b{x} (and vice versa).

answered Mar 21, 2011 at 23:15

cjm

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1 Comment

Walt Stoneburner Over a year ago

Thanks, I've been staring at the square brackets and seeing parenthesis.

Brad Gilbert · Accepted Answer · 2011-03-22 00:33:25Z

You need to read the perlref documentation that talks about references.

There is a difference in how your arrays are stored:

# this is an array in an array variable
@a = (1, 2, 3);

Verses storing a reference to an array:

# this is an array in a scalar variable stored as a reference to the previous array:
$b = \@a;

Functionally that works like a pointer. Thus you can also store this reference in a hash:

$x{'something'} = \@a;

This all works just fine. What you haven't realized is that []s create references to an array, which you can't store in an array variable. You must store it in a scalar. Thus this instead:

$c = [1, 2, 3];
$x{'else'} = $c;

will work.

And accessing and modifying the array before you do the second assignment can be done using:

$c->[3] = '4';

or by using it in array form dereferencing it first

push @$c, 5;

Will Budic · Accepted Answer · 2020-11-18 18:50:51Z

0

Your example helped me with hashes. To end up in my code with:

# Following is confusing as hell. We look to store in the hash an array reference.
# But must convert back and fort via an scalar, since actual arrays returned 
# from an hash are copies in perl. i.e. my @a = $lists{$e}; will not work.

       my $a = $lists{$e};
       if(!$a){$a=();$lists{$e} = \@{$a};}
       push @{$a}, $v;

Thank you very much. Will

answered Nov 18, 2020 at 18:50

Will Budic

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