Once you understand how references work, there's no surprise. \@say is the reference to @say, when you change the contents of @say, \@say still points to @say, i.e. to the new contents of it. When you store a reference, you don't store the contents.

> when I dump \@say, it prints out the expected results (different for each animal)

Of course. Otherwise, changing @say would have to edit the already printed output.

($q=q:Sq=~/;[c](.)(.)/;chr(-||-|5+lengthSq)`"S|oS2"`map{chr |+ord
}map{substrSq`S_+|`|}3E|-|`7**2-3:)=~y+S|`+$1,++print+eval$q,q,a,
[download]

Hello sdani,

I would say it's a quite unexpected behaviour..., because when I dump \@say, it prints out the expected results...

That’s because Data::Dumper (like Data::Dump) applies a dereferencing operation to its argument if that argument is a reference. This is usually what you want; but to avoid it, just print the variable instead:

23:27 >perl -MData::Dumper -wE "my @a = qw( a e i o u ); print Dumper(
+\@a); print \@a;"
$VAR1 = [
          'a',
          'e',
          'i',
          'o',
          'u'
        ];
ARRAY(0x4d2ec8)
23:27 >
[download]

Note the square brackets in the output from Data::Dumper: they denote an anonymous array reference. See perlreftut.

Hope that helps,

You are changing the inhabitants of always the same (hotel) room and every time storing the address to this one room.

Your dumps inside the loop only show the temporary inhabitants, but your last dump will only show the last inhabitants for always the same address.

Using my will tell Perl to always create a new proper room with a unique address.

Clearer?