I still cannot correctly grasp the
where syntax in Julia.
I’d like to know if one can use Julia’s
where keyword as in Haskell; e.g., similar to the following Haskell snippet:
len x y = sqrt (sq x + sq y) where sq a = a * a
the Julia code for which could be
len(x, y) = sqrt( sq(x) + sq(y) ) where sq(a) = a * a end
Please let me know if this is (or will be) possible.