How does Ampersand pass arguments into #to_proc as "obj"? —receiving_method(&:method)

Question

receiver_method(&:method) functionality seems clear enough, yet I find a gap in the explanations* about the assignment to 'obj' as in—

class Symbol
  def to_proc #<parameter notably absent>
    Proc.new { |obj, *args|
      obj.send (self, *args)
    }
  end
end

How does this assign the receiver method's object/passed argument as 'obj'?

* What does map(&:name) mean in Ruby?

* https://www.skorks.com/2013/04/ruby-ampersand-parameter-demystified/

score 1 · Accepted Answer · answered Oct 13 '18 at 11:42

How does Ampersand pass arguments into #to_proc as “obj”? — receiving_method(&:method)

Ampersand does not pass anything anywhere. Ampersand converts the argument to a proc instance, implicitly calling to_proc on it. And passes the result as a block to the caller.

Let’s stick with the example:

%w[42 foo].map(&:to_i)
#⇒ [42, 0]

What’s going on here?

• to_i is being converted to proc as you shown in the OP

#                            ⇓⇓⇓⇓⇓
proc { |obj, *args| obj.send(:to_i, *args) }

• we pass this proc to the caller (without lose of generality, I’d write it with block syntax for the sake of clarity

%w[42 foo].map do |obj, *args|
  obj.send(:to_i, *args)
end

• NB! *args is off-side here, since map passes the single argument to the block:

%w[42 foo].map do |obj|
  obj.send(:to_i)
end

That would map:

'42' → '42'.send(:to_i) ≡ '42'.to_i → 42,
'foo' → 'foo'.send(:to_i) ≡ 'foo'.to_i → 0,

yielding:

#⇒ [42, 0]

the receiver method* actually calls the proc (not yields-to), I gather? the question I first-linked, says 'uses as the method's block' *a specific type of method known as a caller? — kruchkamgar, Oct 13 '18 at 19:21

How does Ampersand pass arguments into #to_proc as "obj"? —receiving_method(&:method)

1 Answers1