@array3 = keys %{{ map {$_ =>1} @array1, @array2 }};
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Oooooh!

The force is strong in that one.

Rob

@array3 = keys %{{ map {$_ =>1} @array1, @array2 }};

I see no need for a map here:

my @union = keys %{{(@array1,@array2)x2}};
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AM said:
I see no need for a map here:

my @union = keys %{{(@array1,@array2)x2}};
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That doesn't work when (@array1+@array2) is even (unless you don't mind losing some of your data :)

my @array1=qw(a b);
my @array2=qw(c d);
my @union = keys %{{(@array1,@array2)x2}};
print "$_\n" for @union;

#prints:
a
c
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my %hash = ();
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@hash{ @array1, @array2 } = ();
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This is a list assignment; that is, this single line
assigns into multiple keys. Just as $hash{$foo} = 0
creates a string key from $foo in the hash %hash,
so @hash{ @array } creates a bunch of keys, whose
members are named in @array, in the hash %hash. Also,
since two arrays mashed together form one big array--
that is, (@a1,@a2,@a3) becomes one flat array-- so then
the assignment @hash{ @array1, @array2 } creates string
keys out of each element in the arrays for %hash.
Since we know that keys are by nature unique, the
hashtable has removed the duplicates for us, for free!
So, now all there is left to do is extract the keys
from the hashtable:

@array3 = keys %hash;
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...and we're done.

Rob

Provide a function that takes two arrayrefs, combines them and returns an array containing all the elements in the previous two arrays, bu no duplicates.

As I don't know them, you can't use commandline opts. *grins*

my @arr1 = (1, 2, 3, 4, 5);
my @arr2 = (4, 5, 6, 7, 8);

my @arr3 = f(\@arr1, \@arr2);

sub f {
#234567890#234567890#234567890
@h{map{@$_}@_}=1;keys%h
}
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Update: 22 chars

@h{map@$_,@_}=1;keys%h
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------
We are the carpenters and bricklayers of the Information Age.

Don't go borrowing trouble. For programmers, this means Worry only about what you need to implement.

20 chars:

sub union {
keys%{{map{$_,1}@_}}
}
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- danboo

EDIT: oops, didn't pay attention to instructions. scratch that.