(contributed by brian d foy)
Calling a subroutine as C<&foo> with no trailing parentheses ignores
-the prototype of C<foo> and passes it the current value of the argumet
+the prototype of C<foo> and passes it the current value of the argument
list, C<@_>. Here's an example; the C<bar> subroutine calls C<&foo>,
which prints what its arguments list:
By default, your program starts in package C<main>, so you should
always be in some package unless someone uses the C<package> built-in
with no namespace. See the C<package> entry in L<perlfunc> for the
-details of empty packges.
+details of empty packages.
=head2 How can I comment out a large block of Perl code?