Writing foldr and foldl in elixir

my_list.exs

# Looks like List.foldr and List.foldl are delegated to erlang: 
# https://github.com/elixir-lang/elixir/blob/v1.0.3/lib/elixir/lib/list.ex#L116L118
# But if we wanted to implement these in elixir, seems like we can only write fold1, but not foldr

# Question: Can we define `foldr` without using high-level functions (ie. List.reverse) 
# and using a recursive approach similar to `foldl`?  I really can't think how but would be
# very interested if someone can show me a solution (if it exists).

defmodule MyList do
  def foldl([], acc, _) do
    acc
  end
  
  def foldl([head|tail], acc, func) do
    foldl(tail, func.(head, acc), func)
  end
 
  def foldr() do
  end
end
 
ExUnit.start
 
defmodule MyListTest do
  alias MyList, as: L
  use ExUnit.Case
  
  test "it folds a list from right to left" do
    list = ["e", "l", "i", "x", "i", "r"]
    assert L.foldl(list, "", &(&1 <> &2)) == "rixile"
  end
  
  test "it folds a list from left to right" do
    list = ["e", "l", "i", "x", "i", "r"]
    assert L.foldr(list, "", &(&1 <> &2)) == "elixir"
  end
end

defmodule MyList do
  def foldr([h | []],acc,f), do: f.(h,acc)
  def foldr([h | t],acc,f) do
    f.(h,foldr(t,acc,f))
  end

  def dbg(x,acc) do
    IO.puts x
    x * acc
  end
end

And then to demonstrate we are actually folding from the right 'MyList.foldr([1,2,3,4],1,&MyList.dbg/2)' yields:

4
3
2
1
24