In a method that take multiple optional parameters, how can any but the first be specified?

Question

I have a method like this:

def foo(fruit='apple', cut="sliced", topping="ice cream")
  # some logic here
end

How can I call it where I only override the topping parameter but use the default values for the others, something like this

foo('','','hot fudge')

Of course this does not work as intended, but I want to only provide a value for the third optional parameter, and have the first two stick with their default values. I know how to do this with a hash, but is their a shortcut way to do it, using the above syntax?

Answer 1

As of Ruby 2.0, you can use keyword arguments:

def foo(fruit: 'apple', cut: "sliced", topping: "ice cream")
  [fruit, cut, topping]
end

foo(topping: 'hot fudge') # => ['apple', 'sliced', 'hot fudge']

Answer 2

You can't use this syntax to do this in ruby. I would recommend the hash syntax for this.

def foo(args={})
  args[:fruit]    ||= 'apple'
  args[:cut]      ||= 'sliced'
  args[:topping]  ||= 'ice cream'
  # some logic here
end

foo(:topping => 'hot fudge')

You could also do this using positional arguments:

def foo(fruit=nil,cut=nil,topping=nil)
  fruit    ||= 'apple'
  cut      ||= 'sliced'
  topping  ||= 'ice cream'
  # some logic here
end

foo(nil,nil,'hot fudge')

Bear in mind that both of these techniques prevent you from passing actual nil arguments to functions (when you might want to sometimes)

Answer 3

No. You have to check the value of the parameters inside function foo. If they are the empty string, or null, you can set them to the default parameter.