That would probably match if not beat join, if you weren't stringifying the array twice.
Actually you did that previously and I carelessly cargo-culted it into my code.
However upon testing it's actually very slightly faster! I don't know why:
perl -MBenchmark=:all -MAlgorithm::Combinatorics=:all -wle'
for ("1".."4") {
my $t0 = Benchmark->new;
my $i = variations_with_repetition(["a".."z"],$ARGV[0]);
my @x; push @x, pack "(A*)*", @$_ while $_=$i->next;
my $t1 = Benchmark->new;
my $ts = timestr(timediff($t1,$t0));
print $ts }' 4
1 wallclock secs ( 0.77 usr + 0.01 sys = 0.78 CPU)
1 wallclock secs ( 0.77 usr + 0.01 sys = 0.78 CPU)
1 wallclock secs ( 0.75 usr + 0.00 sys = 0.75 CPU)
1 wallclock secs ( 0.75 usr + 0.01 sys = 0.76 CPU)
perl -MBenchmark=:all -MAlgorithm::Combinatorics=:all -wle'
for ("1".."4") {
my $t0 = Benchmark->new;
my $i = variations_with_repetition(["a".."z"],$ARGV[0]);
my @x; push @x, pack "A*", qq[@$_] while $_=$i->next;
my $t1 = Benchmark->new;
my $ts = timestr(timediff($t1,$t0));
print $ts }' 4
0 wallclock secs ( 0.74 usr + 0.01 sys = 0.75 CPU)
1 wallclock secs ( 0.72 usr + 0.01 sys = 0.73 CPU)
1 wallclock secs ( 0.73 usr + 0.00 sys = 0.73 CPU)
1 wallclock secs ( 0.74 usr + 0.01 sys = 0.75 CPU)
perl -MBenchmark=:all -MAlgorithm::Combinatorics=:all -wle'
for ("1".."4") {
my $t0 = Benchmark->new;
my $i = variations_with_repetition(["a".."z"],$ARGV[0]);
my @x; push @x, join "", @$_ while $_=$i->next;
my $t1 = Benchmark->new;
my $ts = timestr(timediff($t1,$t0));
print $ts }' 4
1 wallclock secs ( 0.65 usr + 0.01 sys = 0.66 CPU)
1 wallclock secs ( 0.64 usr + 0.00 sys = 0.64 CPU)
0 wallclock secs ( 0.63 usr + 0.00 sys = 0.63 CPU)
1 wallclock secs ( 0.64 usr + 0.01 sys = 0.65 CPU)
BTW, that is one seriously quick machine you're running. What is it?
MacBook Pro i7
3.1GHz (4.1GHz Turbo Boost) 16GB RAM 1TB SSD
Darwin 10.13.4
STOP REINVENTING WHEELS, START BUILDING SPACE ROCKETS!—CPAN 🐪
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