Can Anybody Please Explain (my $self = Shift) In Perl

Answer :

First off, a subroutine isn't passed the @ARGV array. Rather all the parameters passed to a subroutine are flattened into a single list represented by @_ inside the subroutine. The @ARGV array is available at the top-level of your script, containing the command line arguments passed to you script.

Now, in Perl, when you call a method on an object, the object is implicitly passed as a parameter to the method.

If you ignore inheritance,

 $obj->doCoolStuff($a, $b); 

is equivalent to

 doCoolStuff($obj, $a, $b); 

Which means the contents of @_ in the method doCoolStuff would be: @_ = ($obj, $a, $b);

Now, the shift builtin function, without any parameters, shifts an element out of the default array variable @_. In this case, that would be $obj.

So when you do $self = shift, you are effectively saying $self = $obj.

I also hope this explains how to pass other parameters to a method via the -> notation. Continuing the example I've stated above, this would be like:

sub doCoolStuff {   # Remember @_ = ($obj, $a, $b)   my $self = shift;   my ($a, $b) = @_; 

Additionally, while Moose is a great object layer for Perl, it doesn't take away from the requirement that you need to initialize the $self yourself in each method. Always remember this. While language like C++ and Java initialize the object reference this implicitly, in Perl you need to do it explicitly for every method you write.

In top level-code, shift() is short for shift(@ARGV). @ARGV contains the command-line arguments.

In a sub, shift() is short for shift(@_). @_ contains the sub's arguments.

So my $self = shift; is grabbing the sub's first argument. When calling a method, the invocant (what's left of the ->) is passed as the first parameter. In other words,

$o->method(@a) 

is similar to

my $sub = $o->can('method'); $sub->($o, @a); 

In that example, my $self = shift; will assign $o to $self.

If you call:

$myinstance->myMethod("my_parameter");   

is the same that doing:

myMethod($myinstance, "my_parameter");   

but if you do:

myMethod("my_parameter");   

only "my_parameter" wil be passed.

THEN if inside myMethod always you do :

 $self = shift @_;   

$self will be the object reference when myMethod id called from an object context
but will be "my_parameter" when called from another method inside on a procedural way.
Be aware of this;