> $ perl -le '$u=1; ($y=$u*=5)++; print $y' > 6 It's interesting to note that this parse (due to precedence) as ($y=($u*=5))++, not (($y=$u)*=5)++
This is important for overloaded operators (which are going to become much easier to do in Perl6). The importance arises if Perl6 allows assignment to be overloaded like Ruby does. If it does, then the two expressions aren't guaranteed to be identical as they are now in Perl5. Rob
