Perl 5 is smart about copying lists. It only copies as many items as are on the left hand side. It works because list assignment in scalar context yields the number of items on the right hand side. So, n
items will be created by the regex, but they won't be copied and discarded, just discarded. You can see the difference the extra copy makes in the naive case in the benchmark below.
As for efficiency, an iterative solution is often easier on memory and CPU usage, but this must be weighed against the succinctness of the goatse secret operator. Here are the results of benchmarking the various solutions:
naive: 10
iterative: 10
goatse: 10
for 0 items:
Rate iterative goatse naive
iterative 4365983/s -- -7% -12%
goatse 4711803/s 8% -- -5%
naive 4962920/s 14% 5% --
for 1 items:
Rate naive goatse iterative
naive 749594/s -- -32% -69%
goatse 1103081/s 47% -- -55%
iterative 2457599/s 228% 123% --
for 10 items:
Rate naive goatse iterative
naive 85418/s -- -33% -82%
goatse 127999/s 50% -- -74%
iterative 486652/s 470% 280% --
for 100 items:
Rate naive goatse iterative
naive 9309/s -- -31% -83%
goatse 13524/s 45% -- -76%
iterative 55854/s 500% 313% --
for 1000 items:
Rate naive goatse iterative
naive 1018/s -- -31% -82%
goatse 1478/s 45% -- -75%
iterative 5802/s 470% 293% --
for 10000 items:
Rate naive goatse iterative
naive 101/s -- -31% -82%
goatse 146/s 45% -- -75%
iterative 575/s 470% 293% --
Here is the code that generated it:
#!/usr/bin/perl
use strict;
use warnings;
use Benchmark;
my $s = "a" x 10;
my %subs = (
naive => sub {
my @matches = $s =~ /a/g;
return scalar @matches;
},
goatse => sub {
my $count =()= $s =~ /a/g;
return $count;
},
iterative => sub {
my $count = 0;
$count++ while $s =~ /a/g;
return $count;
},
);
for my $sub (keys %subs) {
print "$sub: @{[$subs{$sub}()]}
";
}
for my $n (0, 1, 10, 100, 1_000, 10_000) {
$s = "a" x $n;
print "
for $n items:
";
Benchmark::cmpthese -1, \%subs;
}
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