> $ 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


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