Luckily in these cases, the faster answer is also the clearest answer in either case.
On Mon, Jan 29, 2018 at 5:32 PM Paul Johnson <p...@pjcj.net> wrote: > 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 >
