当前的Ruby方法是通过super调用的吗？

Question

在运行时的方法中,有没有办法知道是否已super在子类中调用该方法？例如

module SuperDetector
  def via_super?
    # what goes here?
  end
end

class Foo
  include SuperDetector

  def bar
    via_super? ? 'super!' : 'nothing special'
  end
end

class Fu < Foo
  def bar
    super
  end
end

Foo.new.bar # => "nothing special"
Fu.new.bar  # => "super!"

我怎么写via_super?,或者,如果有必要的话via_super?(:bar)？

Answer 1

我的其他答案、@mudasobwa和@sawa 的答案加上递归支持的最终组合：

module SuperDetector
  def self.included(clazz)
    unless clazz.instance_methods.include?(:via_super?)
      clazz.send(:define_method, :via_super?) do
        first_caller_location = caller_locations.first
        calling_method = first_caller_location.base_label

        same_origin = ->(other_location) do
          first_caller_location.lineno == other_location.lineno and
            first_caller_location.absolute_path == other_location.absolute_path
        end

        location_changed = false
        same_name_stack = caller_locations.take_while do |location|
          should_take = location.base_label == calling_method and !location_changed
          location_changed = !same_origin.call(location)
          should_take
        end

        self.kind_of?(clazz) and !same_origin.call(same_name_stack.last)
      end
    end
  end
end

唯一不起作用的情况（AFAIK）是如果您在基类中有间接递归，但我不知道如何通过解析代码来处理它。