Want to protect your cyber security and still get fast solutions? Ask a secure question today.Go Premium

x
  • Status: Solved
  • Priority: Medium
  • Security: Public
  • Views: 253
  • Last Modified:

Passing 2D array to a sub()

Hi,

I read a 2d array from file, so I have something like:

    @my2dArray[5][5];

I can iterate over it, ok. I want to pass it to a sub routine, like:

    pass2dArray(my2dArray);

    sub pass2dArray()
    {
        my (@array) = @_;
    }

Am I passing a copy of the 2d array to the function, or is it just referencing the original array?

Then, if I want to pass a single row of the array to another sub, how would I do that? I was hoping something like:

    sub pass2dArray()
    {
        my (@array) = @_;
   
        # pass row 2:
        passOneRowOf2dArray(2);
       
        # etc
        passOneRowOf2dArray(4);
    }

    sub passOneRowOf2dArray()
    {
        my (@row) = @_;
        for (my $i = 0; $i < @row; $i++) {
           ...
        }
    }
   

Thanks
0
DJ_AM_Juicebox
Asked:
DJ_AM_Juicebox
  • 3
  • 2
1 Solution
 
Adam314Commented:
With this:
    pass2dArray(@my2dArray);
You are passing a copy of the original array, which contains references to the 2nd dimension.  You are not passing copies of the 2nd dimension.  Meaning, if your pass2dArray function, if you add or remove elements from @array, this will not be affected in @my2dArray.  But if you change, add or remove elements from any of the rows, this will be affected in @my2dArray.

To pass one row as a reference, you would use:
    passOneRowOf2dArray($array[2]);    #passes reference to second row
To pass one row as an array (not a reference), you would use:
    passOneRowOf2dArray(@{$array[2]});

If you want to pass a copy of a structure like this, you can use the Storable module's dclone function:
http://search.cpan.org/~ams/Storable-2.18/Storable.pm
0
 
DJ_AM_JuiceboxAuthor Commented:
>>To pass one row as a reference, you would use:
    passOneRowOf2dArray($array[2]);    #passes reference to second row


Ok, that's just what I put, but how do you recover the reference to the single row in the function? Like:

    sub passOneRowOf2dArray()
    {
        my @row = shift;
    }

???

That doesn't work though...

Thanks

0
 
Adam314Commented:

sub passOneRowOf2dArray {
    my $ref_to_row = shift;    #get reference
    
    #changes to $ref_to_row will show up in the original array
    print "first value=$ref_to_row->[0]\n";
    print "all values=" . join(", ", @{$ref_to_row}) . "\n";
    
    #Get as a regular array, makes a copy
    my @array_copy = @{$ref_to_row};
    
    #changes to @array_copy will not show up in the original array
    print "first value=$array_copy[0]\n";
    print "all values=" . join(", ", @array_copy) . "\n";
}

Open in new window

0
 
Adam314Commented:
This website might be helpful:
http://perldoc.perl.org/perlreftut.html
0
 
DJ_AM_JuiceboxAuthor Commented:
Perfect, thanks.
0

Featured Post

Concerto's Cloud Advisory Services

Want to avoid the missteps to gaining all the benefits of the cloud? Learn more about the different assessment options from our Cloud Advisory team.

  • 3
  • 2
Tackle projects and never again get stuck behind a technical roadblock.
Join Now