这种方法如何超过Rubocop账户分支条件大小？

Question

I have a simple before_save method used to assign an account_id based on whether a user_id , application_id or contact_id is already present in the model.

class Note < ApplicationRecord
  belongs_to :contact
  belongs_to :application
  belongs_to :user
  belongs_to :account

  before_save :assign_account_id

  private

  def assign_account_id
    self.account_id =
      if user_id.present?
        user.account.id
      elsif application_id.present?
        application.account.id
      elsif contact_id.present?
        contact.account.id
      end
  end
end

The method works and, in my opinion, is about as simple as I can get it, but Rubocop insists it is slightly over the Assign Branch Condition size (ABC size), where the limit is 15 and my method's ABC is 15.33 .

According to this article , the ABC size of 15 is achieved with 8 assignments, 8 branches and 8 conditions. However, I only count 1 assignment self.account_id = , 1 branch (the return), and three conditions (the 3 if/elsif statements).

Am I mistaken? Where are the additional assignments, branches or conditions coming from? The calls to present? , traversing the model hierarchy?

NOTE: I'm note looking for alternative implementations, I'm interested in understanding what is causing this score.

---

For anyone interested, here's the solution I eventually went with that satisfies the ABC size.

self.account_id = [
  user&.account&.id,
  application&.account&.id,
  contact&.account&.id
].find(&:present?)

I chose it because of the vertical list conveys the cascading nature of the fields the most strongly. I felt I'd be able to return to this and still be able to grok what it's doing.

Answer 1

This is the web page that the rubocop documentation references in its source code , in the documentation for Metrics/AbcSize (as of the latest version; 0.61.0 ).

To paraphrase, it says that:

A scalar ABC size value (or "aggregate magnitude") is computed as: |ABC| = sqrt((A*A)+(B*B)+(C*C)) |ABC| = sqrt((A*A)+(B*B)+(C*C))

Where A is the number of Assignments , B is the number of Branches and C is the number of Conditions .

• Your code has 1 Assignment ( self.account_id = ).
• Your code has 15 Branches (!!!) ( user_id , .present? , user , .account , .id , application_id , .present? , application , .account , .id , contact_id , .present? , contact , .account and .id )
• Your code has 3 Conditions ( if ... elsif ... elsif ).

Plugging this into the above formula gives:

ABC = sqrt(1*1 + 15*15 + 3*3)
    = sqrt(235)
    = 15.32970...

And that's where the (rounded) value of 15.33 is coming from.

---

I know you're not really asking for an alternative implementation, but here's one anyway:

def assign_account_id
  self.account_id = (user || application || contact).account.id
end

...And you could even consider moving those brackets into a separate method.