不同命名空间继承问题：如何获取应用命名空间实例？

I've built a custom gem with the following code:

module GemNamespace
  class Foo; end
  class Bar
    def foo
      @foo ||= Foo.new
    end
  end
end

In my application, I'm subclassing these gem classes under my own namespace:

module ApplicationNamespace
  class Foo < GemNamespace::Foo; end
  class Bar < GemNamespace::Bar; end
end

When I call the foo method on my application's Bar instance, it spits out a GemNamespace::Foo object instead of my desired ApplicationNamespace::Foo:

bar = ApplicationNamespace::Bar.new
puts bar.foo => #<GemNamespace::Foo:0x007f849d8169f0>

I want to get an instance of ApplicationNamespace::Foo instead, without having to override the foo method in my application's Bar class. How can I make this work?

The Fix: Use Dynamic Constant Lookup in the Gem #

The root issue here is that the gem's Bar#foo method is hardcoding Foo.new, which Ruby resolves to GemNamespace::Foo because the method lives inside the GemNamespace module. To make this method play nicely with subclasses, we just need to adjust how it references the Foo class—making it dynamic instead of static.

Update your gem's Bar class to use one of these approaches:

Option 1: Use Class Namespace Resolution

module GemNamespace
  class Foo; end
  class Bar
    def foo
      @foo ||= self.class::Foo.new
    end
  end
end

Option 2: Explicitly Fetch the Constant

module GemNamespace
  class Foo; end
  class Bar
    def foo
      @foo ||= self.class.const_get(:Foo).new
    end
  end
end

Why This Works #

When you create an instance of ApplicationNamespace::Bar, self inside the foo method points to that instance. So self.class returns ApplicationNamespace::Bar.

Ruby's constant lookup rules will then:

1. First check if ApplicationNamespace::Bar has its own Foo class (it doesn't in your case)
2. Traverse up to the enclosing namespace ApplicationNamespace, where it finds your ApplicationNamespace::Foo class

This way, the method automatically uses the correct Foo subclass based on the instance's actual class—no overrides needed in your application code.

内容的提问来源于stack exchange，提问作者kudrin