Question to seek clarification on the definition of "..." in type signatures

[hjwylde]

Hi! I have a question regarding ..., specifically relating to #2540 / #1293

Prior to seeing this, I thought that:

• ... is used to denote overloaded method type signatures in the same class/module
• Overriding a type signature from a class or module overwrites the parent signature, and it is no longer possible to reference it

After seeing this, I realise I possibly didn't understand ..., as it looks to me that:

• ... is also used to denote "defer to the parent type signature", similar to the super keyword in Ruby
• ... only denotes super if there are no real method overloads

Take the following examples:

module Foo
  def call: -> Integer
end

class Example1
  include Foo
  def call: -> Float | ...
end

class Example2
  include Foo
  def call: -> Float | ... 
end
class Example2 # Overload/re-open
  def call: -> Symbol
end

It looks to me (by RBS and Steep) that:

• Example1 uses ... to denote "super", and has type -> Integer | -> Float
• Example2 uses ... to denote "overload", and has type -> Float | -> Symbol

I think that it would be great to hear what the intention of ... is, but I'll also just mention a couple of my thoughts on it:

• It feels problematic to me that a library may wish to use ... to denote super, but then the super functionality can be overridden if a user of the library adds an actual overload of the method signature.
• Re-using ... to denote super could lead to confusion when reading type signatures - it is no longer clear whether this means the type signature is overloaded, or whether it is referring to super, especially if you have to know/understand if there are any re-openings of the class/module.
• I think having a separate syntax for super may be better here for behaviour and readability

Thanks :)