Ruby self, operator precedence and instance method invocation

Question

Working through some code examples and came across this snippet, edited here for brevity:

  class Year
    attr_reader :number

    def initialize(number)
      @number = number
    end

   def a(str)
     puts str
   end

   def %(other)
     number % other # why number instead of @number
   end

   def my_func()
     (self a "hello") # error
   end

   private

   def divisible_by?(i)
     (self % i) == 0 #Operator precedence binding self and % together?
   end
end

I'm looking to understand why:

1. the call to (self % i) doesn't require a space between the self and %, I'm guessing it has something to do with operator precedence?

2. Why does, within the method definition for %(other) the method refer to number, I would've expected it to have to refer to the instance variable @number.

Any help would be appreciated and/or links to ruby docs that would help me explain this. ty in advance

Answer 1

why number instead of @number

Because you have defined a reader/getter, might as well use it. Today method number is backed by an instance variable, tomorrow it's computed (or lazily instantiated, etc.). By using the method and not its internals, you shield yourself from cascading changes. It's called "encapsulation". But you could have used the variable, it's just not a good practice.

the call to (self % i) doesn't require a space between the self and %, I'm guessing it has something to do with operator precedence?

No. Nothing to do with precedence. There was no ambiguity in the spaceless form and ruby was able to parse it successfully, that's why it's allowed.