I agree that this isn't necessarily the best task to use recursion for, but here's one possibility:

use Data::Dumper;

sub mk_chunker
{
    my $n = shift;
    die unless $n > 0;
    my $sub;
    return $sub = sub {
        return \@_ if @_ < $n;
        return (
            [ @_[0..$n-1] ],
            $sub->(@_[$n..$#_]), # recursion
        );
    };
}

*chunk_5 = mk_chunker(5);

print Dumper chunk_5('a' .. 'z');

[download]

The closure in mk_chunker is a good candidate for Perl 5.16's new __SUB__ feature...

use 5.016;

sub mk_chunker
{
    my $n = shift;
    die unless $n > 0;
    return sub {
        return \@_ if @_ < $n;
        return (
            [ @_[0..$n-1] ],
            __SUB__->(@_[$n..$#_]), # recursion
        );
    };
}

[download]
perl -E'sub Monkey::do{say$_,for@_,do{($monkey=[caller(0)]->[3])=~s{::}{ }and$monkey}}"Monkey say"->Monkey::do'

In reply to Re: How to return a list using recursion? by tobyink
in thread How to return a list using recursion? by dr.jekyllandme

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