macro_as_arg.ex

defmodule Foo do
  defmacro foo_macro(i) do
    quote do
      unquote(i)
    end
  end

  def bar(foo_macro(1) = i), do: IO.puts(i)
  def bar(foo_macro(2) = i) do
    IO.puts "This better be 2: #{i}"
  end
end

Foo.bar(1)
# => 1

Foo.bar(2)
# => This better be 2: 2

Foo.bar(3)
#** (FunctionClauseError) no function clause matching in Foo.bar/1
#    foo.exs:8: Foo.bar(3)
#    (elixir) lib/code.ex:370: Code.require_file/2

Hi. The macro-expansion happens before the function is defined, so there's no sense in which the macro is an "argument" to the function. During macro expansion it expands, in this case, into the values passed to it, so that when it's time to compile the function the only thing left is the binding: 1 = i or 2 = i. As long as the code macros expand to is valid for the compiler, you can use macros anywhere. I hope this helps. If it doesn't, let me know and maybe I can help in some other way.

@wobh yeah, that was my position, but others at the meeting didn't think it works work that way. This is a small POC to show them it actually worked.