On Fri, 1 Nov 2002, Ed Peschko wrote: > I'm probably opening up a whole new can of worms here, but if we said > that the following were both vector operators: > > ^ == intersection operator > v == union operator > > then these could have potentially useful meanings on their *own* as set > operators, as well as modifying other operators. For example: > > @a = (1,2,3); > @b = (4,1,3); > > @a = @a ^ @b; # @a = (1,3); > @a = @a v @b; # @a = (1,2,3,4);
Or is @a = (1,2,3,4,1,3) ? So what happens with: @a = (1, 2, 3, 3); @b = (4, 3, 1, 3); @a = @a ^ @b; # @a = (1,3) ? @a = (1, 3, 3) ? element order? @a = @a v @b, # @a = (1, 2, 3, 4) ? @a = (1, 2, 3, 4, 4, 4, 1, 3) ? Does the operand order change something? The element order is significant in arrays, but not in hashes. And with hashes, each key is unique. For a hash, these operators should be simpler. %a = ( 1 => 'a', 2 => 'b', 3 => 'c' ); %b = ( 4 => 'A', 1 => 'B', 3 => 'C' ); And the operand order could be used to apply a "keep the values from the left-hand side hash" rule: %a = %a ^ %b; # %a = (1 => 'a', 3 => 'c') %a = %b ^ %a; # %a = (1 => 'B', 3 => 'C') %a = %a v %b; # %a = (1 => 'a', 2 => 'b', 3 => 'c', 4 => 'A') %a = %b v %a; # %a = (1 => 'B', 2 => 'b', 3 => 'C', 4 => 'A') (And notice how I avoided answering my own questions about arrays by swiftly speaking of hashes... ;-) -- Philippe "BooK" Bruhat Few things in life are so dependable as the incompetence of those who know precisely what they are doing. (Moral from Groo The Wanderer #5 (Pacific))
