--- Peter Cornelius <[EMAIL PROTECTED]> wrote:
> This might be a little more clear if you break down
> the way arguments
> are being passed in and what they actually are. It
> sounds like
> you're aware that the arguments are passed in as a
> list name @_, it
> looks like the arguments are something like this:
>
> @_ = ( #<---- A list
> { _name => 'something' }, # <---- a hash
> reference (objects are
> hashes in perl)
> 'some set name string' #<---- a string
> );
>
> The first version extracts the arguments so that you
> can refer to
> them by names that might have some meaning to
> someone maintaining
> your code latter on. The second version accesses
> them directly.
>
> > sub name{ #version 1
> > my $self=shift;
>
> shift is going to give you $_[0]
>
> >
> > my $set_name=shift;
>
> This sets $set_name to $_[1]
>
> >
> > $self->{_name}=$set_name if defined $set_name;
> >
>
> So this is equivalent to
> $_[0]->{_name} = $_[1] if defined $_[1]
> the '->' is to dereference the hash reference stored
> in $_[0].
>
> > return $self->{_name}
> > }
> >
> >
> > Another version for the subroutine name
> >
> > sub name{ #verstion 2
> > $_[0]->{_name}=$_[1] if defined }$_[1];
> > {$_[0]->{_name}
> > }
> >
> > I feel a little bit confuse about the verion 2
> > subroutine name. The way it gets the value of
> > attribute name looks like this to me:
> > array element->{_name}
> >
> > (I know that the $_[0] in the default array @_ is
> > actually an object $self. )
>
> Hope this helps
> PC
Thank you all for the reply.
Based on what I learn the regular method to defer a
hash reference to get specific value takes this
format:
$ref_hash->{key1}
but in this line
$_[0]->{_name}= $_[1] if defined $_[1]
the format is
array element->{_name}
Is the middle man $ref_hash is omitted in this format?
Does this what Perl really sees:
$_[0]=$ref_hash;
$ref_hash->{_name};
and put these two lines into one line to make it
short:
$_[0]->{_name}
Li
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