How does Ruby’s .() operator work?

The dot-parentheses notation is a shorthand way for passing arguments to the implicit call method on a Ruby object:

foo = lambda {|bar| puts bar}

foo.call('baz')
#=> baz
foo.('baz')

foo.call('baz') === foo.('baz')
#=> true

Also note that the following notations are also valid (and equivalent) invocations of the call method:

foo['baz']
#=> baz
foo::('baz')
#=> baz

In your example, you’re explicitly overriding the call method on the TestServiceObject class such that it returns the string 'method' when called. Accordingly, you can explicitly override the the call method to accept arguments:

class TestServiceObject
  def call(foo=nil)
    foo || 'method'
  end
end

TestServiceObject.new.()
#=> method
TestServicesObject.new.('bar')
#=> bar

UPDATE:

As commenter @LoganSerman duly notes, the shorthand operator appears to work on anything that responds to call, which is validated in part by the following example:

m = 12.method("+")

m.call(3)
#=> 15
m.(3)
#=> 15

UPDATE 2:

As commenter @Stefan also points out from the documentation on Proc#call:

prc.() invokes prc.call() with the parameters given. It’s a syntax sugar to hide “call”.