recursion - Explanation for group by 3 elements from a list function using FOLD in OCAML

Question

I have a piece of code that does the following: group 3 elements of a list of n elements. The main function is called group_by_3. For example, executing group_by_3 [1;2;3;4;5;6;7] will give me ([1;2;3],[4;5;6],[7]).

let group_by_3 lst =
  let accum = ( [], [], 0 )
  in
  let f (all_groups, current_group, size) x =
    if size = 3
    then ( (List.rev current_group) :: all_groups, 
           [x], 1 )
    else ( all_groups, x::current_group, size+1)
  in
  let (groups, last, _) = List.fold_left f accum lst
  in List.rev ( List.rev last :: groups)

I don't really understand why this works (it is provided in class).

• What are all_groups, current_group, size?

• What does this do?

  if size = 3
  then ( (List.rev current_group) :: all_groups, 
         [x], 1 )
  else ( all_groups, x::current_group, size+1)
  

Thank you!

Answer 1

All groups is the final answer, current is the group of three elements filled while parsing the input and size is the number of elements actually in current I guess.

Do you know the behavior of fold_left?