Re: Trying to Understand Callback(s)

[shmem]

...but I don't really quite get the idea of callbacks, and particularly I don't understand the syntax.

Well, it's not just quite like with a phone - you ring up somebody, they hang up and call you - but rather a way to parametrize a subroutines execution chain via a subroutine reference (like a function pointer). The called sub may have been compiled into any context, and the compilation rules of that context apply. And what ikegami said.

I don't really get the first line of the get_callback routine either, when nothing appears to be being passed across

It gets something passed across but discards it.

A sub routine called as a method ($thingy->routine(@args)) by something ($thingy - irrelevant whether $thingy is a literal, a blessed reference, a scalar holding a string, or an expression which resolves to a package or a blessed reference) gets passed the invocant as its first parameter in @_.

So, having this preamble

sub routine { 
   my($thingy, $tx) = @_; # $thingy is not used
   print "routine: $tx\n";
}

my $ref = \&routine; # take a reference to the sub
my $pkg = "main";
my $obj = bless do { \my $x };

# we could also say...
#   my $pkg = __PACKAGE__;
# ...to get any package we are in in compile time

the following call expressions are all equivalent:

# method calls

main->routine("foo");
main->$ref   ("foo");
$pkg->routine("foo");
$pkg->$ref   ("foo");
$obj->routine("foo");
$obj->$ref   ("foo");
# or even
(join'',map{chr}(109,97,105,110))->routine("foo");
__PACKAGE__->$ref("foo");
(bless \my $x)->$ref("foo");

# function calls

routine ('main', "foo");
$ref -> ('main', "foo");
routine ($pkg,   "foo");
$ref -> ($pkg,   "foo");
&$ref   ($pkg,   "foo");

routine ($obj,   "foo");
# etc.

More on differences between subroutine calls either as function or method: perlsub, perlmod, perlref.

The expression

my (undef, $thing) = @_;

is just a fancy way of saying

shift @_; # drop first element of argument list
my ($thing) = @_;

The subroutine get_callback you quoted just seems not to care at all about it's invocant, but nonetheless expects to be called as a method. Hmm, code smell...

[DanielSpaniel]

Thank you all so much for your replies. I sincerely appreciate your time and efforts in making things clearer for me. I'll need to re-read a couple of times at least, but I have at least a much better understanding now. Thank you again.

[Laurent_R]

Well, it is a bit difficult to understand if you don't practice. I think you really need to start using these things in your code and make your own experiments to really grasp them. Once you get a working knowledge on them, you'll probably wonder how you could live without them before.

If you're interesting in digging into these techniques further, please read this excellent book: https://hop.perl.plover.com/book/ by dominus. You can read it for free on-line, but I should warn you that you might end up willing to buy the paper copy. At least, that's what happen to me when I first read it about ten years ago.