actually:

*@x=(1,2,3,4);*
*$z=*x;*
*print "$z\n";*
*print "@$z\n";*
*print %$z."\n";*

produced:

**main::x*
*1 2 3 4*
*0*

which now baffles me even more :)

On Sun, Apr 17, 2011 at 5:06 PM, Issac Goldstand <[email protected]>wrote:

>  I'm not sure, I understand.
>
> Let's take your example one step further and say we had values for @x and
> %x
>
> Given the same code, if you tried to get @$y and %$y, you (probably)
> wouldn't get anything useful.
>
> But if you tried to get @$z or %$z, you should get back @x and %x
>
>   Yitzchak
>
>
>
> On 17/04/2011 17:02, Ronen Angluster wrote:
>
> a bit,
> if i got this right, $z=*x is assigning the singleton(of sorts) of
> "main::x" into $z?
>
> On Sun, Apr 17, 2011 at 4:58 PM, Issac Goldstand <[email protected]>wrote:
>
>>  Hi Ronen!
>>
>> The former (\$x) is a reference to the scalar value of x ($x).  The latter
>> is a pointer to the symbol table for all symbols called "x".
>>
>> So when you're doing $\$x ($$y) you're asking for the scalar value of the
>> reference to the scalar 1
>> When you're doing $*x ($$z) you're asking for the scalar entry in the
>> symbol table for symbol x (also 1).
>>
>> Does that clear things up for you?
>>
>>   Yitzchak
>>
>>
>> On 17/04/2011 16:52, Ronen Angluster wrote:
>>
>>  Hi all!
>> happy passover!
>>
>>  i've been wondering and could not find anyone who could explain
>> to me what is the difference between \$var to *var...
>> consider the following:
>> *$x=1;*
>> *$z=\$x;*
>> *$y=*x;*
>> *print "$y\n$z\n";*
>> would print:
>>  **main::x*
>> *SCALAR(0x1951ac8)*
>>
>>  but if we:* print "$$y\n$$z\n";*
>> the output would be:
>> *1*
>> *1*
>>
>>  so, why use 1 against the other? what does it mean that the pointer is
>> to the class and
>> not to the memory space of the variable?
>>
>>
>>  Ronen
>>
>>
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