Apparently I wrote it badly, but my question concerning scalar vs. $#array was
whether there was a difference in run-time costs, not the functional difference.
I don't have a specific issue, but code can always run faster. The question is
whether I should use scalar() without concern, or whether I should try to cache
the result. For instance, if scalar is O(n), then this loop:
for ( my $i = 0 ; i < scalar(@objs) ; ++i ) { ... }
becomes O(n^2) instead of O(n). That's exactly the kind of thing that adds up
without raising an obvious red flag in profiling.
One could then make it O(n) by doing
my $limit = scalar(@objs);
for ( my $i = 0 ; $i < $limit; ++$i ) { ... }
If $# is O(1) instead of O(n) then it would be good to have the habit of using
for (my $i = 0 ; $i <= $#objs ; ++$i) { ... }
Or, if one is building a list and needs to check later if the list is
non-empty, should one set a flag during the list build or just use scalar() at
the end?
The question is, should I try to make a habit of one of these in particular? Or
does it matter?
For question (2), I tried option (C) but its failure was masked by other
problems. Once I had those fixed, I realized that it didn't work. I thought
about changing it to \( map ... ) but that would seem to be clearly inferior to
[ map ... ].
At 04:11 AM 4/10/2006, Brian Raven wrote:
>Alan M. Carroll <> wrote:
>> I went netsearching for this but couldn't find anything in all the
>> noise.
>>
>> 1) What is the runtime cost of the scalar operator? Is it constant or
>> does it depend on the size of the array? Is it different than
>> $#array?
>
>Evaluating an array in a scalar context returns the size of the array,
>$#array returns the index of the last element. So, not the same.
>
>As for the cost, why does it matter if your code is fast enough?
>Seriously, it might help answer your question if we knew what problem
>you are actually trying to solve.
>
>C doesn't do what you want, so it gets disqualified from the contest and
>retires to the changing rooms in shame for an early shower. Did you try
>it, BTW?
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