Is the default operator a macro?

Questions / Help
1

HexDoc on default args

This states that the compiler creates new functions given the default operator. Is this actually a macro?

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2

\\ is parsed as an operator (like + or == or -> etc) but not normally defined anywhere:

iex(1)> 4 \\ 5
** (CompileError) iex:1: undefined function \\/2 (there is no such import)

iex(1)> quote do
...(1)>   4 \\ 5
...(1)> end
{:\\, [], [4, 5]}

The implementation of def transforms the AST that’s passed to it with pattern matching, but no function named \\ ever runs:

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3

No, look that you cannot use \\ in function body. Elixir is capable of parsing a predefined set of operators. Some of them are macros, but some of them are used only within other macros (like defor defp) and because of that their AST representation is used instead.

(…) these operators appear in the precedence table above but are only meaningful within certain constructs:

Source: Operators reference — Elixir v1.16.0

Take a look at a simplest example:

defmodule Example do
  defmacro sample({:\\, _, [left, right]}) do
    quote bind_quoted: [left: left, right: right] do
      if is_nil(left) do
        {:right, right}
      else
        {:left, left}
      end
    end
  end
end

iex> require Example
Example
iex> Example.sample(1 \\ 2)
{:left, 1}
iex> Example.sample(nil \\ 2)
{:right, 2}

Since \\ is a valid operator you are also able to define it, so it can be used inside function body:

defmodule Example do
  def left \\ right do
    {left, right}
  end
end

iex> import Example
Example
iex> 1 \\ 2  
{1, 2}
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Thank you. I have a lot to learn.