Assigning anonymous hash to a list

[Joseph Werner]

Hmmm.  Yitzchak Scott-Thoennes, I recognize, Abigail I do not (perhaps
I should?)

So, Yitzchak, is this, in fact, a simple matter of operator precedence
in that the ',' operator is of lower precedence than the assignment
operator (absurd though that may seem)  AND not a case of a scalar
value (which is the result of a Perl expression involving a ','
operator) being offered to a  list assignment?

Always interested in a learning opportunity,
Christian

On Tue, Jul 30, 2013 at 5:43 PM, Yitzchak Scott-Thoennes
<sthoenna at gmail.com> wrote:
> On Tue, Jul 30, 2013 at 2:32 PM, Joseph Werner <telcodev at gmail.com> wrote:
>> On Tue, Jul 30, 2013 at 4:51 PM, Abigail <abigail at abigail.be> wrote:
>>> By that argument, this is a scalar assignment as well:
>>>
>>>   my ($i1, $i2, $i3) = (4, 5, 6);
>>
>> No, What you have done here is to assigned a list value to an array of
>> assignable elements.
>
> There is *no* array anywhere there.  There are two lists, though.
>
>> I am talking about the example at the top of this thread, which was a
>> scalar assignment to a list of elements.
>
> I can see what you mean when you say that, but it is not correct terminology.
>
> Perl has two assignment operations, list assignment, and scalar assignment.
> Which it is is determined purely based on what is on the left of the assignment.



-- 
Best Regards,
[Joseph] Christian Werner Sr
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