After all, &lc is how you tell Perl that you want
to call a function named "lc", as opposed to the default behavior
attached to that name ... namely, the lc operator.
I didn't try &lc. I tried \&lc. (Is the backslash showing up?)
I tried to take the reference of lc, because I thought (mistakenly)
that built-ins and subroutines could be treated the same way.
Yes, the backslash came through. But taking a ref to a sub
only works if there's a sub there to take a ref to. The idea
of \&lc reminds me of that famous description of
Oakland: ``There's no `there' there.''