Dear All,

I have been learning how to write non-blocking http request.
While following the example from Mojo::UserAgent:
http://mojolicious.org/perldoc/Mojolicious/Guides/Cookbook#Non-blocking

I do not understand the meaning of the following:
$fetch->() for 1 .. 2;
The following is the complete example:

use Mojo::UserAgent;
use Mojo::IOLoop;

my @urls = (
  'mojolicious.org/perldoc/Mojo/DOM',  'mojolicious.org/perldoc/Mojo',
  'mojolicious.org/perldoc/Mojo/File', 'mojolicious.org/perldoc/Mojo/U
+RL'
);

my $ua = Mojo::UserAgent->new(max_redirects => 5);
$ua->transactor->name('MyParallelCrawler 1.0');

my $delay = Mojo::IOLoop->delay;
my $fetch;
$fetch = sub {

  return unless my $url = shift @urls;

  my $end = $delay->begin;
  $ua->get($url => sub {
    my ($ua, $tx) = @_;
    say "$url: ", $tx->result->dom->at('title')->text;
    $end->();

    $fetch->();
  });
};

#Process two requests at a time
$fetch->() for 1 .. 2;
$delay->wait;
[download]

I have tried the following:
1) Removing the "for 1 .. 2" part.
i.e. $fetch->();
the result is the http request will always get the first link only.

2) Increase the end point, say "for 1 .. 4"
Based on the comment from the provided example, I believe this will result in processing 4 requests at a time.
Could I ask your help in explaining the meaning of the above syntax?
Does those 1, 2 become the argument passing into the function? (but $fetch is not reading @_, so this shouldn't be the case)
What exactly does that for loop mean in this context?


Many thanks for your help in sharing your Perl knowledge,
Ronald

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