Question: Default argument example explanation about elixir-in-action HOT 2 CLOSED

Yep! Perfect sense thank you. It doesn't cover both /3 options, and d will be the default value of 2 unless you supply all four arguments, because elixir matches to the left. I have a Ruby background so this is in line with my assumptions. Thanks for your help!

from elixir-in-action.

Good question! The best way to be sure is to verify it ;) Let's change the function to print each args:

defmodule MyModule do
  def fun(a, b \\ 1, c, d \\ 2) do
    IO.puts "a = #{a}"
    IO.puts "b = #{b}"
    IO.puts "c = #{c}"
    IO.puts "d = #{d}"
  end
end

$ iex test.exs

iex> MyModule.fun(:arg_1, :arg_2, :arg_3)
a = arg_1
b = arg_2
c = arg_3
d = 2

If you think about it, it makes sense. If a single default arg is omitted when calling a fun, then the right-most arg with a default (in our case the arg d) will have the default value.

Does that make sense?

from elixir-in-action.