Even if you have a sub q{}, calling q() will be parsed as the q() operator. Calling &q() or main::q() gets you the function. This test verifies this behavior for nine different operators.

from irc://irc.perl.org/p5p 2004/08/12

<kane-xs>  bug or feature?
<purl>     You decide!!!!
<kane-xs>  [kane@coke ~]$ perlc -le'sub y{1};y(1)'
<kane-xs>  Transliteration replacement not terminated at -e line 1.
<Nicholas> bug I think
<kane-xs>  i'll perlbug
<rgs>      feature
<kane-xs>  smiles at rgs
<kane-xs>  done
<rgs>      will be closed at not a bug,
<rgs>      like the previous reports of this one
<Nicholas> feature being first class and second class keywords?
<rgs>      you have similar ones with q, qq, qr, qx, tr, s and m
<rgs>      one could say 1st class keywords, yes
<rgs>      and I forgot qw
<kane-xs>  hmm silly...
<Nicholas> it's acutally operators, isn't it?
<Nicholas> as in you can't call a subroutine with the same name as an
           operator unless you have the & ?
<kane-xs>  or fqpn (fully qualified package name)
<kane-xs>  main::y() works just fine
<kane-xs>  as does &y; but not y()
<Andy>     If that's a feature, then let's write a test that it continues
           to work like that.