No. Because normal behaviour is order-insensitive, I expect that f(a=1,b=2) and f(b=2,a=1) should have identical behaviour. If you make the arguments order-sen

No. Because normal behaviour is order-insensitive, I expect that f(a=1,b=2) and f(b=2,a=1) should have identical behaviour. If you make the arguments order-sensitive, then they may not. If, for readability reasons, I change the order of the arguments, I will then be surprised when my code stops working. Conversely, If (because I think it is order-sensitive) I believe that, in order to make the function do what I want, I have to call f(a=1,b=2) and not f(b=2,a=1), then I'm not going to change the order willy-nilly, so I won't be affected by my expectations not matching reality.