cblavier

cblavier

Hey there :waving_hand:

I have the 2 following structs.

defmodule Variation do
  @enforce_keys [:id, :attributes]
  defstruct [:id, :description, :attributes, :slots, :block]
end

defmodule VariationGroup do
  @enforce_keys [:id, :variations]
  defstruct [:id, :description, :variations]
end

I would like to pattern match both structs at once, based on the fact they share similar attributes: :id and :description.

Is this achievable through behaviors?

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JohnnyCurran

JohnnyCurran

Just treat them as maps:

def my_func(%{id: id, attributes: attributes}) do
  # work
end

And if you need to access the underlying module:

def my_func(%{id: id, attributes: attributes} = entity) do
  entity.__struct__
end
fuelen

fuelen

Another options:

def test(%module{}) when module in [Variation, VariationGroup], do: ...
def test(term) when is_struct(term, Variation) or is_struct(term, VariationGroup), do: ...
cblavier

cblavier OP

I will actually use both of your solutions.

Guard + map attribute pattern matching

Thanks!

rvirding

rvirding

Creator of Erlang

Don’t you still have to know which one you have when you work with them?

cblavier

cblavier OP

I have a general purpose loop that doesn’t need to know which one is which.

Then I also have specific functions that precisely pattern match against either %Variation{} or %VariationGroup{}

cblavier

cblavier OP

I just realized that what I’m looking after is polymorphism for structs.
But I don’t think this is available in Elixir. Is it?

michallepicki

michallepicki

You can define a Protocol and have separate implementation for each struct

cblavier

cblavier OP

But what if I also want an implementation that apply on both structs?

michallepicki

michallepicki

If you want to share the implementation of a function from a protocol, you can move it to a different module and delegate there from all separate protocol implementations. Or you can create a separate protocol for the shared parts and pass a list of struct modules (or primitive types if applicable) to defimpl.

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