Haskell CPS Programming Style Problem

Question

here is a function multiplies the elements in a list using CPS style

mlist xx k = aux xx k
  where aux [] nk = nk 1
    aux (0:xs) nk = k 0
    aux (x:xs) nk = aux xs $ \v -> mul x v nk

what if I change the 'k' to 'nk' in the expression aux (0:xs) nk = k 0, whats the difference between the two ?

Answer 1

k is always the original continuation passed to mlist whereas for the list [1, 0] nk in that case will be \v -> mul 1 v k (from third case of aux).

If we assume mul is defined as mul x y k = k $ x*y, this doesn't make a practical difference since y will always be 0. But the actual method of arriving at that result is different (barring possible optimizations by the compiler).

Answer 2

The difference is that the original definition "short circuits" any built-up multiplications passed by tail call applications, while the changed expression only short-circuits testing the values - it keeps the built-up "version" of the continuation function.