«

$#array
scalar @array or
my $n = @array ?

On 08/04/11 21:30, George Mpouras wrote:
> $#array
> scalar @array or
> my $n = @array ?

perldoc Benchmark

Sounds like premature optimization, to me.

On 2011-08-04 23:30, George Mpouras wrote:

> which is faster ?
>
> $#array
> scalar @array or
> my $n = @array ?

If magic or tie or operator-overload is not involved, there will hardly
be any difference in speed. Just benchmark to find out.

$#array is the highest index of @array. The other two are about the
number of elements. $[ gives the minimum index which is normally 0.

--
Ruud

George Mpouras <> wrote:
>$#array
>scalar @array or
>my $n = @array ?

IMNSHO you are asking the wrong question. Because each of these three
code pieces has a different semantic the right question would be "Which
code piece has the right semantic for my algorithm?"

jue

On 04-Aug-2011 3:30 PM, George Mpouras wrote:
> $#array
> scalar @array or
> my $n = @array ?
I assume with strict the my one with work faster, I think the strict
pragma makes it faster perl.
-r

robin <> wrote:
>On 04-Aug-2011 3:30 PM, George Mpouras wrote:
>> $#array
>> scalar @array or
>> my $n = @array ?
>I assume with strict the my one with work faster, I think the strict
>pragma makes it faster perl.

Your statement can be interpreted in two ways:
- I assume with strict the my one with work faster than the other two
options shown
- I assume with strict the my one with work faster than without using
strict

Which one is it?

jue

robin <> writes:
> On 04-Aug-2011 3:30 PM, George Mpouras wrote:
>> $#array
>> scalar @array or
>> my $n = @array ?
> I assume with strict the my one with work faster, I think the strict
> pragma makes it faster perl.

According to the corresponding documentation, access to 'my' variables
should be faster than acceses to 'other variables' and I expect this
to be actually true because a variable declared with my reside in a
scope-specific array and are accessed by array index while 'package
globals' require a hash lookup in the symbol table hash of the package
they belong to. But that's not at all related to 'strict' which
reconfigures the Perl compile by modifying the variable $^H and has no
runtime effects. (AFAIK).