[erlang-questions] Can I do the same with a fold easily

[Jachym Holecek]

# Roelof Wobben 2015-10-18:
> I have this two functions :
> 
> filter(List, Filter) ->
>     lists:filter(Filter, List).

What is the point of this?

> split2(List) ->
>     { filter(List, fun(X) -> math_functions:odd(X) end),
> filter(List, fun(X) -> math_functions:even(X) end) }.
> 
> which produces both {[1,3],[2,4]}
> 
> Could I this also do with a fold or is this the best solution.

Entirely untested, but the best is to 1) keep it real 2) keep it simple. :-)

    foo(L) ->
        foo(L, [], []).

    foo([X | L], Es, Os) ->
        case (X rem 2) of
            0 ->
                foo(L, [X | Es], Os);
            1 ->
                foo(L, Es, [X | Os])
        end;
    foo([], Es, Os) ->
        {Es, Os}.

Add reverse/1 calls to the terminal branch if you particularly care about the
order of results.

BR,
	-- Jachym