Since we are on the subject, am I right in thinking that the 'last index' will always be one less than scalar @array? Or is there something more to it?

Almost always true...

If it wasn't for the fact that you can change the base-index of perl arrays to something other then 0... But I've never seen anyone use that feature yet. The global parameter for the feature is '$['.

So use $# to get the last defined element index so you can safely add (or loop), and use scalar @array to find out how many elements there are in the array...

Read perldoc perldata again!

Version 5 of Perl changed the semantics of $[: files that don't set the value of $[ no longer need to worry about whether another file changed its value. (In other words, use of $[ is deprecated.) So in general you can assume that
scalar(@whatever) == $#whatever + 1;

For getting the 'last defined index' in an array - if you are using it as a subscript - I prefer $array[-1] = $foo; over $array[$#array] = $foo;

$#array is still useful if you are looping over an array by index, like this

for my $i (0..$#array) {
  # do something with $array[$i]
}
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-- Hofmator

You are correct...

Once again, perldoc perldata to the rescue...

scalar(@whatever) == $#whatever + 1;