ERLANG - splitting lists into lists

Question

Hi, this is my first post, I hope that you are doing well. So I'm just starting erlang, and I have a problem, I'm not sure how to handle it.

So I have a binary that I get in the form

<<56, 23, 67, 34, 45, 78, 01, 54, 67, 87, 45, 53, 01, 34, 56, 78>>

My goal is to split it into an optional list (or binary, if more efficient) based on 01.

For example, the above should look like this:

<<56, 23, 67, 34, 45, 78>> <<54, 67, 87, 45, 53>> <<34, 56, 78>>

-or -

[[56, 23, 67, 34, 45, 78], [54, 67, 87, 45, 53], [34, 56, 78]]

01 is a separation tag; it does not need to be included in the final output.

I tried something like this: (PLEASE neglect if there is a better way)

parse1([]) -> [];
parse1(1) -> io:format("SOHSOHSOHSOHSOHSSOHSOHS");
parse1(Reply) -> parse1({Reply, []});
parse1({Reply, nxtParse}) ->
    [H | T] = Reply,
    case H of
         _ when H > 1 ->
            [H | nxtParse],
              io:format("Reply 1 = ~p~n", [H]),
            parse1({T, nxtParse});
         _ when H == 1 -> 
            io:format("SOHSOHSOHSOHSOHSSOHSOHS");

        [] ->
            ok

    end.

This is not at all clean, and it is not at all like what I am writing about. I am sure that I slammed my head "spirit" when someone tells me.

, , . , ERL BIF -, , .

. -B

Answer 1

R14A, Erlang binary :

1> Bin = <<56, 23, 67, 34, 45, 78, 01, 54, 67, 87, 45, 53, 01, 34, 56, 78>>.
<<56,23,67,34,45,78,1,54,67,87,45,53,1,34,56,78>>
2> binary:split(Bin, <<01>>, [global]).
[<<56,23,67,34,45,78>>,<<"6CW-5">>,<<"\"8N">>]

, (<<"6CW-5">> <<"\"8N">>), , - , , . . :

3> io:format("~w~n", [binary:split(Bin, <<01>>, [global])]).
[<<56,23,67,34,45,78>>,<<54,67,87,45,53>>,<<34,56,78>>]
ok