Reverse an integer regardless of its sign

Thank you very much for clarifying it.

Originally I noticed this on other data too, but I found that it is not a problem in this specific case, because as far as I know Integer.digits/1 will never return list like [1, -2, 3].

There is nothing surprising with 83 as result, because we have: (3 * 10^0) + (-2 * 10^1) + (1 * 10^2) i.e. (3 * 1) + (-2 * 10) + (1 * 100) i.e. 3 - 20 + 100 which gives 83.

Hi @NobbZ
Thanks so much for sharing.

 -123_456_789
    |> Integer.digits() # => [-1, -2, -3, -4, -5, -6, -7, -8, -9]
    |> Enum.reverse() # => [-9, -8, -7, -6, -5, -4, -3, -2, -1]
    |> Integer.undigits() # => -987654321

But:

Integer.undigits([1, -2, 3]) # => 83

This seems a very good topic of conversation.

Ah right, my bad - I have edited my first post

Hello @Ted

, Super nice!
I didn’t know that was possible. I’m learning so much, thanks for sharing.

Why converting to a string at all?

Just do the maths:

defmodule DR do
  def r(n), do: r(n, 0)

  def r(0, x), do: x
  def r(n, x), do: r(div(n, 10), x * 10 + rem(n, 10))
end

Hello @LostKobrakai

Super nice patter matching idea

Yes!

I’m learning so much, thanks for sharing.

Wow, that is clean!

I’m curious about the reason for two function heads because it seems like the first head works for both positive and negative:

iex(7)> 123 |> Integer.digits() |> Enum.reverse() |> Integer.undigits()    
321
iex(8)> -123 |> Integer.digits() |> Enum.reverse() |> Integer.undigits()
-321

@Andres: I can see much simpler way:

defmodule Example do
  def sample(integer) when is_integer(integer) do
    integer |> Integer.digits() |> Enum.reverse() |> Integer.undigits()
  end
end

Is that what you wanted?

Edit: Code changed after @Ted comment