does it mean that the following should print 1 (but it doesn't)? perl -le 'x(@y);sub x{$_[0]=1;print @y}'

perl -le 'x($y[1]);sub x{$_[0]=1;print $c++,": $_" for @y}'
0: 
1: 1
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In your version calling x(@y) flattens the (empty) array @y into an empty list. There's no element of @y to be aliased and passed within @_ into x().

Note that indexing an array does not create slots within:

perl -le '$c = $y[10]; x($y[1]);sub x{$_[0]=1;print $c++,": $_" for @y
+}'
0: 
1: 1
[download]

The array @y isn't 11 elements long merely because I looked at index 10. The sentence you asked about describes that for subroutines. An argument to a subroutine can be a nonexistent element of an array or hash (not an array or hash), and it only springs into existence if told so via its alias inside the sub (by assigning or referencing).

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}