BTW what is the third argument($i) nFor( $n-1, @_, $i ) doing here?

As the sub recurses, one of the array references is remove from the front of @_ at each level, and the current value being iterated by the for loop at that level is added to the end.

Once $n == 0, all the array references have been removed and $n has been shifted off, all that is left in @_, is the set of elements to be printed.

BTW. Here is a cleaner implementation that takes a callback to which the results sets are passed:

#! perl -slw
use strict;

sub nForX(&@) {
    my $code = shift;
    my $n = shift;
    return $code->( @_ ) unless $n;
    for my $i ( @{ shift() } ) {
        &nForX( $code, $n-1, @_, $i );
    }
}

my @a = 1..10;
my @b = 'a'..'z';
my @c = map chr, 33 .. 47;

nForX {
    print join ' ', @_;
} 3, \( @a, @b, @c );
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thanks a lot. yes i get it now. In the same light.....i want to call nfor() with different number of arguments...

nfor(1,\@a_1); 
nfor(2,\@a_1,\@a_2);
nfor(3,\@a_1,\@a_2,\@a_3);...so on 
nfor(N,\@a_1,\@a_2,...\@a_N).
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I have a way...but what to check if there is a much crispier way to do it.

If you store your arrays (more specifically references to your arrays) in an array, you can say

my @arrayOfArrays = ( \@a_1,\@a_2,...,\@a_20 );
my $n = 10;
nfor( $n, @arrayOfArrays[ 0..$n-1 ] );
[download]

Where will you get the variable number of arrays from? And how will you be storing them?

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