You wouldn't have this problem if you were using $var[n] instead of $varn.

my @arrays = map { [ split /;/, $_ ] } @nums;
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But to answer your question:

my $num1 = "1;2;3";
my $num2 = "4;5;6";

my (@ar1, @ar2);

for (
   [ $num1, \@ar1 ],
   [ $num2, \@ar2 ],
} {
   my ($num, $ar) = @$_;
   @$ar = split /;/, $num;
}
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By the way, don't fool yourself into thinking the first argument of split is a text string. split expects a regexp.

@ar1 = split(';', $num1);
@ar2 = split(';', $num2);
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Update: if you want to know how to get

# given: @num = ("1;2;3", "4;5;6");
$array[0] = [1, 2, 3];
$array[1] = [4, 5, 6];
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@array = map { [ split(';', $_) ] } (@num);
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(@ar1[0..2], @ar2[0..2]) = map { split(';', $_) } ($num1, $num2);
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Perl flattens arrays. List assignment is greedy.

The sooner you stop trying to fight it and just accept these fact, the happier your Perl programming experience will be :-)

--


See the Copyright notice on my home node.

"The first rule of Perl club is you do not talk about Perl club." -- Chip Salzenberg

not quite understand why dose Perl flatten the results.

Because Perl functions can only return one thing: a list of scalars. A list of scalars is created from the two arrays and returned by map.

Then you ask to assign that list to a pair of arrays. There's no way for Perl to know how to split that list to assign it to two arrays, so it assigns the entire list to the first array.