Define functions in module metaprogrammically

I have three identical methods in my module, that do (almost) exactly the same thing. Instead of repeating the function definitions, I am trying to define them once to keep code minimal and same for all of them.

So far I’ve tried using Code.eval_string:

defmodule MyModule do
  Enum.each ~w(method1 method2 method3), fn method ->
    @method method

    Code.eval_string """
      def #{@method}(map) when is_map(map) do
        do_map_stuff(:#{@method}, map)
      end

      def #{@method}(arg) do
        do_other_stuff(:#{@method}, arg)
      end
    """
  end

  ## Other Methods

end

but this throws ArgumentError:

Compiling 1 file (.ex)
** (ArgumentError) cannot invoke def/2 outside module
    (elixir) lib/kernel.ex:4297: Kernel.assert_module_scope/3
    (elixir) lib/kernel.ex:3299: Kernel.define/4
    (elixir) expanding macro: Kernel.def/2
    nofile:1: (file)

I think quote/unquote might be the way to go, but I’m not exactly sure how to do this using them (I’ve already read the Meta Guide on the Elixir website).

2 Likes

Don’t need to use Code.eval_string; you can wrap a def in a “loop” and use unquote/1 to “extract” the variable into the context that you’re def-ing in:

defmodule M do
  Enum.each ~w(foo bar), fn(method_name) ->
    def unquote(:"#{method_name}")(), do: unquote(method_name)
  end
end
3 Likes

There’s no need to use eval_string, you could use Code.eval_quoted but it’d be much much better to just define a macro in another module and then import it.

Here’s a quick example, you can put on a file foo.ex and execute with elixir.

defmodule Gen do

  # Note that macro arguments and return values are quoted expressions
  defmacro gen({name, _, _}) do
    quote do

      # here we define it dinamically
      def unquote(name)() do
        IO.inspect(unquote(name))
      end

    end
  end
end

defmodule Foo do
  import Gen
  gen foo
end

Foo.foo
1 Like

Thank you. Someone just posted this as an answer on my SO question as well.