> : @costs = (sort { $a.key <=> $b.key } > : { amortize($^_) => $^_ }^.(@cost ^* $inflation) > : )^.value; > > I wonder how the parser is going to tell that the start of the second > line is supposed to be an anon sub, rather than an anon hash. It looks > to me as if it'll require substantial lookahead, which could become > a problem.
I suspect it will actually require *less* lookahead than Perl 5 would. As soon as it encounters the $^_ it knows it's seeing a higher-order function. Damian