Question regarding Mojo::UserAgent non-blocking example

[Ronald Wong]

Hi All,

I have recently started learning how to write non-blocking http request and Mojo::UserAgent is the best one which can help me to overcome the proxy issue within the company environment.
While learning by following the example, there is something I do not understand from the code and raised a question in perlmonks:
http://www.perlmonks.org/?node_id=1202496;showspoiler=1202547-1

However, I am still clueless regarding the behaviour, could I ask your help in explaining it further?

http://mojolicious.org/perldoc/Mojolicious/Guides/Cookbook#Non-blocking

The following is the code 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;


From the above example, at the last 2nd line, "$fetch->() for 1 .. 2;",
if I remove the for loop, the code will process the first URL only.

I suppose the $fetch is a recursion and the function calling itself within the function.
Shouldn't the code process 1 request each time if I remove the for loop?

I am wondering inside the $fetch function that $fetch->() should go before $end->()?

As following:
$fetch = sub {
...
    $fetch->();

    $end->();
  });
};

instead of 
$fetch = sub {
...
    $end->();

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


After the above change, the code will now process all the URLs one by one when it is only $fetch->().

And if I run it with for-loop like $fetch->() for 1 .. 2, it looks like processing 2 URLs in 1 go.
The code behaving much more like the original intended.

The problem with the original code is that if you are calling it 1 time only, at the time reaching the end of the event loop the condition may have satisfied and exit the program before it can call another function itself.(my guess) 
Am I correct with the above theory?


Thanks,
Ronald

[Sebastian Riedel]

> $fetch = sub {
> ...
> $fetch->();
>
> $end->();
> });
> };

I think you are correct, this code is better.

--
Sebastian Riedel
http://mojolicio.us
http://github.com/kraih
http://twitter.com/kraih

[Ronald Wong]

Many thanks for your reply and especially thanks for creating such great module!