Your making an assumption that turns out to be wrong :-)

First:

sub y { \&x(); }
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Returns a reference to whatever x() returns *) but functions can not return plain arrays; they returns lists or single (scalar) values, depending on how they're called (this is called context in perl).

update: the following is wrong, since \ forces list-context on its argument

If x() would return a list, y() would return a list of references to those values, but that's not what is happening here.

Since you're doing

my $h = &y;
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y() is called in scalar (in stead of list) context. the context is propagated to the returning statement in y() - that's \&x();. That propagates the scalar context up to the my @a = (1,2,3) statement. That statement is a list assignment, and a list assignment in scalar context returns the number of values assigned.

So, x() returns 3, and then you take a reference to that and return it from y() and assign it to $h.

Then you try to derefence $h as an array reference, but since $h contains a reference to a scalar, that won't work.

update: the rest is correct, though

The quickest fix would be:

sub y { [ x() ]; }
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Which creates a new anynomous array containing all the values returned by x() in list context.

*) Personally I think using \&sub(); is really confusing, since \⊂ without parens returns a reference to the subroutine; generally I think it's better to never use &sub(); for calling subroutines, since ⊂ can also cause confusion - see perlsub. In essence, you only need & for taking a reference to a sub, or other special cases, and you probably should use sub(); ( or sub; ) for all "normal" cases.