Numerals in Positional Notation   [12]

Using an accumulating parameter

Just as we did for the evaluation problem, we can resolve this issue of reversed digits by using an accumulating parameter: that is, rather than building up with a cons of each element onto a recursive result, returning nil in the end, we send an extra parameter into each call, build up by consing onto it, and then returning it in the end.

str b 0 = []
str b n = r : str b m
          where (m,r) = n `divMod` b

> str 10 1234
[4,3,2,1]
> str 2 13
[1,0,1,1]

str b a 0 = a
str b a n = str b (r:a) m
            where (m,r) = n `divMod` b

> str 10 [] 1234
[1,2,3,4]
> str 2 [] 13
[1,1,0,1]

Notes

• in order to get the accumulating parameter started off, we need to supply a base case of [] in the beginning: in practice, we could make a main definition which always supplied this for us, and make the recursive definition above just a "helper function"