On Sun, Jan 28, 2018 at 10:57:25PM +0000, Chas. Owens wrote: > $#array is the index of the last element of @array, so it will be one less > than scalar @array which is the number of elements in @array (since Perl > arrays start with index 0). Therefore, to get the number of elements, you > would need to add one to $#array, which makes scalar @array faster. To get > the last index you would need to say @array - 1, which would make $#array > faster.
But it doesn't matter, because this is not a bottleneck in your code. So code for clarity. > On Sun, Jan 28, 2018, 13:39 Peng Yu <pengyu...@gmail.com> wrote: > > > Hi, > > > > For the following two expressions, are they of the same speed or one > > of them is faster? > > > > `$#array` vs `scalar @array` -- Paul Johnson - p...@pjcj.net http://www.pjcj.net -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
