the local @_ is implicitly passed

It's even the same @_. This means that &foo(@_) and &foo; won't behave the same; see perlfaq7, "What's the difference between calling a function as &foo and foo()?". Demonstration:

use strict;

sub foo { pop }

sub bar {
    print "bar before: @_";
    print '&foo      : ' . &foo . "     <--- I am a theif";
    print "bar after : @_   <--- no c";
}

sub baz {
    print "baz before: @_";
    print '&foo(@_)  : ' . &foo(@_);
    print "baz after : @_";
}

bar(qw/ a b c /); print '';
baz(qw/ a b c /);

__END__
bar before: a b c
&foo      : c     <--- I am a theif
bar after : a b   <--- no c

baz before: a b c
&foo(@_)  : c
baz after : a b c
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ihb