On May 27, 2006, at 3:56 PM, chen li wrote:
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}
Yes, the contents of the array element is a hash ref. You could
rewrite this to be the equivalent
${$_[0]}->{_name} = $_[1] if defined $_[1]
Using the '{}' around the $_[0] to more clearly mark it as a reference.
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:
$_[0]->{_name}
It's not really omitted, rather the argument passed in was a hash
reference so the first element of the array ($_[0]) is a hash
reference. You could alias it by saying
$ref_hash = $_[0];
or, if you're feeling confident use it without the alias, as in this
example.
I guess this hash reference is being implicitly passed in by the
method call as part of Perl's OOP implementation so you never do see
the actual parameter passage of
name($ref_hash, $new_name)
Is this what's confusing you?
Hope this helps,
PC
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