SML list option recusion; how to use recursion to output a SOME list

Question

Currently, my code takes in a string s and a string list sl and returns a string list where s was removed (once)

fun all_except_option (s, sl) =
    case sl of
    [] => []
      | hd::tl = if same_string(s, hd)
          then tl
          else hd::all_except_option(s, tl) 

However, what I want is to return NONE if the string s is not in the list and if it is, return SOME (output of the current function). However, I can't simply add SOME( before hd::all_except_option(s, tl) since hd will be appended to something that outputs an option and I can't figure how to do it.

Edit: thanks everyone!

Answer 1

fun all_except_option (s, ss) =
    case ss of
      [] => NONE
    | s'::ss' =>
        if s = s' then SOME ss' else
        case all_except_option (s, ss') of
          NONE => NONE
        | SOME ss'' => SOME (s'::ss')

Note that this only removes the first occurrence of s, which mirrors your version.

You can also use Option.map to avoid the nested case:

fun all_except_option (s, ss) =
    case ss of
      [] => NONE
    | s'::ss' =>
        if s = s' then SOME ss'
        else Option.map (fn ss' => s'::ss') (all_except_option (s, ss'))