Dave Mitchell wrote: > In the current implementation the first two cases are equivalent; > however, in the last (lval) case they differ; with this: > > \(@a) = \($a, $b); \(@a) = \($c, $d); > > @a[0,1] are left aliased to $c and $d, as you'd expect; with this: > > (\(@a) = \($a, $b)) = \($c, $d); > > @a[0,1] are left aliased to $a and $b. I would consider that a bug.Thread Previous | Thread Next