On Sat Sep 18 10:55:52 2010, moritz wrote: > Basic tests added to S03-operators/short-circuit.t. > > 19:52 < moritz_> rakudo: my $x; $x &&= 5; say $x > 19:52 <+p6eval> rakudo a204ba: OUTPUT«5» > 19:52 < moritz_> somebody please confirm that this is wrong > ... > 19:53 < jnthn> moritz_: Looks wrong to me.
Rakudo is correct here -- see S03:3979: "If you apply an assignment operator to a container containing a type object (which is undefined), [...] the operation is defined in terms of the corresponding reduction operator, where the type object autovivifies to the operator's identity value." The identity value for [&&] is True, so Rakudo has this correct. By way of comparison, consider: "my $x; $x *= 5;" which leaves $x with the value of 5. Pm
