rm-rf-etc
Did we get private modules?
I see there was a long discussion on the proposal of private modules, but I can’t tell what decision was reached. I’m writing a library which has an internal-only __using__/1 macro, and I want to know how I can make it private to the library.
Marked As Solved
mudasobwa
To hard-restrict calling macros from allowed places only, one might use __CALLER__ special form inside macro.
@allowed_callers Foo.Bar, Foo.Baz
defmacro any_macro do
raise unless __CALLER__.module in @allowed_callers
...
end
Also Liked
mudasobwa
There is no such thing as “submodule” or “child module” in Elixir. Foo.Bar, despite it shares the first part of the name with Foo is absolutely not related to Foo by any means.
If you meant “to check whether a module shares a namespace,” that’d be easy to achieve with Module.split/1 and List.starts_with?/2, like
{m1, m2} = {Foo.Bar.Baz, Foo.Bar}
[m2, m1] |> Enum.map(&Module.split/1) |> Enum.reduce(&List.starts_with?/2)
#⇒ true
[m1, m2] |> Enum.map(&Module.split/1) |> Enum.reduce(&List.starts_with?/2)
#⇒ false
Nicd
kokolegorille
I made the change to the topic and set the solution as requested.
rm-rf-etc
Love this solution. I would mark it as the solution in this thread but I can’t apparently because of the discussion type I chose.
One question, is there a way to check if __CALLER__.module is a submodule or child module of another module – i.e. ModA.ModB is child of ModA?








