When you understand that Perl arrays cannot hold other arrays, but rather references to arrays, the solutions naturally start occurring. But thinking of them as "multidimensional arrays" is akin to Kirk claiming that Kahn was "thinking in two dimensions" during the final battle of Wrath of Kahn. In other words, by not seeing it as it is, you misunderstand the possibilities.

-- Randal L. Schwartz, Perl hacker

-----------------------------------------------------
Dr. Michael K. Neylon - mneylon-pm@masemware.com || "You've left the lens cap of your mind on again, Pinky" - The Brain
"I can see my house from here!"
It's not what you know, but knowing how to find it if you don't know that's important

How about recursion? You iterate the first dimension, calling yourself with its arrayref.

Almost. A call ->next would return an array of the current 'state' of the system:

(1,a,A)
(1,a,B)
(1,b,A)
(1,b,B)
(2,a,A)
(2,a,B)
(2,b,A)
(2,b,B)

If you want the list generated at once, you could use the following snippet by merlyn. If you want an iterative solution, you could use the following. A benefit being that it should work in relatively constant time no matter how big the arrays are.


my @aoa = (
    [1,2],
    ['a','b','c','d'],
    ['A','B','C'],
);


my $iter = make_aoa_iterative(@aoa);


while (my @els = $iter->() ){
    print @els,"\n";
}


sub make_aoa_iterative {
    my @arefs = @_;
    my @arrayindexes = ();
    foreach (@arefs){
        push @arrayindexes,[$_,0,$#{$_}];
    }
    
    return sub {
            return if $arrayindexes[0]->[1] > $arrayindexes[0]->[2]; 
            my @els =  map { $_->[0]->[ $_->[1]] } @arrayindexes;
            # Check for out of bounds....
            $arrayindexes[$#arrayindexes]->[1]++;
            for (my $i = $#arrayindexes; $i > 0; $i--){
                    if ($arrayindexes[$i]->[1] > $arrayindexes[$i]->[2
+]){
                        $arrayindexes[$i]->[1] = 0;
                        $arrayindexes[$i-1]->[1]++;
                    }else{
                        last;
                    }
            }
            return @els;
            
    };
}
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Update
I'm formal-math challenged. Is this formally a permutation? If someone could clarify, I'd appreciate it.

-Lee

"To be civilized is to deny one's nature."