in reply to referencing list

Just as a note: using $a and $b as variables outside of the sort() function is generally a bad idea.

To answer your question, yes it is possible, but you don't need references for that:

my @x = ( 1, 2, undef ); $x[2] = $x[0] + $x[1]; print @x, "\n";

Or is that not what you are looking for?

~Thomas~
confess( "I offer no guarantees on my code." );