in reply to Re: A hash of lists
in thread A hash of lists

Even though it isn't exactly the same as \@test, for your purposes it works the same.

It's doubtful that his real purposes are anything like the example he showed, so the differences between [ @test ] and \@test may actually be quite significant.

The first, [ @test ] actually creates a new array, a copy of @test, and returns a reference to it. The second, \@test returns a reference to @test itself.

If @test is big, making copies of it could hurt performance.

Also, because the data is copied, changing an element in the array referred to by [ @test ] won't have any affect on the contents of @test itself. Sometimes that's just what you want. Sometimes it isn't.

-sauoq
"My two cents aren't worth a dime.";

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Re: Re: Re: A hash of lists
by wolis (Scribe) on Nov 06, 2003 at 22:42 UTC
    Ah.. now this is interesting..

    In this instance I plan to read lines from a file and split them by their delimiter and store them as lists in a hash.

    So @test in my example is infact one line read from a file, then @test becomes another line etc..

    So what happens to the data if its references all the way down?

    I suspect I should be using the @test method in this case. 

     ___
/\__\ "What is the world coming to?"
\/__/ www.wolispace.com

    [download]
[reply]
[d/l]
      If you iterate over the file, using @test, then [@test] is indeed your only option, as anything else would have your hash continually referencing the same line (the last one). Hope that helps.
[reply]
[d/l]
        Depends on if my is used, doesn't it?

        I went to check, and wrote this script: 

        #!/your/perl/here
use strict;
use warnings;

use Data::Dumper;

my @array = 0..4;
print "\@array ", \@array, "\n";
print Dumper @array;

my %hash;
$hash{one} = \@array;
print "\%hash x1\n";
print Dumper %hash;

my @array = 'a'..'e';
print "\@array ", \@array, "\n";
print Dumper @array;

$hash{two} = \@array;
print "\%hash x2\n";
print Dumper %hash;

print "";

        [download]
        Which gives this output: 
        c:\perl\perl>hash_array.pl
"my" variable @array masks earlier declaration in same 
scope at C:\perl\perl\hash_array.pl line 16.
@array ARRAY(0x1a452dc)
$VAR1 = 0;
$VAR2 = 1;
$VAR3 = 2;
$VAR4 = 3;
$VAR5 = 4;
%hash x1
$VAR1 = 'one';
$VAR2 = [
          0,
          1,
          2,
          3,
          4
        ];
@array ARRAY(0x1ae73c8)
$VAR1 = 'a';
$VAR2 = 'b';
$VAR3 = 'c';
$VAR4 = 'd';
$VAR5 = 'e';
%hash x2
$VAR1 = 'one';
$VAR2 = [
          0,
          1,
          2,
          3,
          4
        ];
$VAR3 = 'two';
$VAR4 = [
          'a',
          'b',
          'c',
          'd',
          'e'
        ];

        [download]
        which differs from the debug x output at the end: 
        c:\perl\perl>perl -d hash_array.pl

Loading DB routines from perl5db.pl version 1.19
Editor support available.

Enter h or `h h' for help, or `perldoc perldebug' for more help.

main::(hash_array.pl:7):        my @array = 0..4;
  DB<1> c 24
[snip Dumper output]
main::(hash_array.pl:24):       print "";
  DB<2> x \%hash
0  HASH(0x1c23ca4)
   'one' => ARRAY(0x1c1b9ac)
      0  0
      1  1
      2  2
      3  3
      4  4
   'two' => ARRAY(0x1c1b9ac)
      -> REUSED_ADDRESS

        [download]
        But if I enclose the last my @array... in a block, the debugger gives the same answer (and the compiler warning goes away).

        Any hints?

        -QM
        --
        Quantum Mechanics: The dreams stuff is made of

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