David Nicol wrote: > what's the advantage of $coderef->&* over $coderef->() ??? $foo->&* means &$foo, so: \$foo->&* # \&$foo $foo->&* # &$foo -- sub call sharing @_ goto $foo->&*; > Also, how about postfix reference operator? > > my $scalar; > my $reference = $scalar -> \*; # instead of \$scalar Then $x = $x->[-1] becomes $x ->= [1->-*].Thread Previous | Thread Next