Optional arguments with default value in Ruby

AdamNYC picture AdamNYC · May 1, 2013 · Viewed 24.6k times · Source

I would like to create a function that has optional arguments with default values

def my_function(a = nil, b=nil, c=500)

end

and call the function with the arguments I would like to specify only

my_function(b=100)

How do I accomplish this in Ruby 1.9.2?

Answer

Jörg W Mittag picture Jörg W Mittag · May 2, 2013

Arguments are bound to parameters like this:

  1. As long as there are unbound mandatory parameters at the beginning of the parameter list, bind arguments left-to-right
  2. As long as there are unbound mandatory parameters at the end of the parameter list, bind arguments right-to-left
  3. Any leftover arguments are bound to optional parameters left-to-right
  4. Any leftover arguments are collected into an array and bound to the splat argument
  5. A block is wrapped up into a Proc and bound to the block argument
  6. If there are any unbound parameters or leftover arguments, raise an ArgumentError

Here's an example:

def foo(mand1, mand2, opt1=:opt1, opt2=:opt2, *splat, mand3, mand4, &block)
  p local_variables.map {|v| "#{v} = #{eval(v.to_s)}" }
end

foo 1, 2, 3
# ArgumentError: wrong number of arguments (3 for 4+)

foo 1, 2, 3, 4
# mand1 = 1
# mand2 = 2
# opt1 = opt1
# opt2 = opt2
# splat = []
# mand3 = 3
# mand4 = 4
# block = 

foo 1, 2, 3, 4, 5
# mand1 = 1
# mand2 = 2
# opt1 = 3
# opt2 = opt2
# splat = []
# mand3 = 4
# mand4 = 5
# block = 

foo 1, 2, 3, 4, 5, 6
# mand1 = 1
# mand2 = 2
# opt1 = 3
# opt2 = 4
# splat = []
# mand3 = 5
# mand4 = 6
# block = 

foo 1, 2, 3, 4, 5, 6, 7
# mand1 = 1
# mand2 = 2
# opt1 = 3
# opt2 = 4
# splat = [5]
# mand3 = 6
# mand4 = 7
# block = 

foo 1, 2, 3, 4, 5, 6, 7, 8 do end
# mand1 = 1
# mand2 = 2
# opt1 = 3
# opt2 = 4
# splat = [5, 6]
# mand3 = 7
# mand4 = 8
# block = #<Proc:0x007fdc732cb468@(pry):42>

So, as you can see both from step 3 above and from the example, you cannot do this, because optional parameters are bound left-to-right, but you want to specify the middle argument.

Note that this has implications on API design: you should design your parameter lists in such a way that the most "unstable" optional parameters, i.e. the ones that a user most likely wants to supply themselves, are furthest to the left.

Ruby 2.0 now has keyword arguments, which is exactly what you are looking for:

def foo(m1, m2, o1=:o1, o2=:o2, *s, m3, m4, key1: :key1, key2: :key2, **keys, &b)
  puts local_variables.map {|v| "#{v} = #{eval(v.to_s)}" }
end

foo 1, 2, 3
# ArgumentError: wrong number of arguments (3 for 4+)

foo 1, 2, 3, 4
# m1 = 1
# m2 = 2
# o1 = o1
# o2 = o2
# s = []
# m3 = 3
# m4 = 4
# key1 = key1
# key2 = key2
# b = 
# keys = {}

foo 1, 2, 3, 4, 5
# m1 = 1
# m2 = 2
# o1 = 3
# o2 = o2
# s = []
# m3 = 4
# m4 = 5
# key1 = key1
# key2 = key2
# b = 
# keys = {}

foo 1, 2, 3, 4, 5, 6
# m1 = 1
# m2 = 2
# o1 = 3
# o2 = 4
# s = []
# m3 = 5
# m4 = 6
# key1 = key1
# key2 = key2
# b = 
# keys = {}

foo 1, 2, 3, 4, 5, 6, 7
# m1 = 1
# m2 = 2
# o1 = 3
# o2 = 4
# s = [5]
# m3 = 6
# m4 = 7
# key1 = key1
# key2 = key2
# b = 
# keys = {}

foo 1, 2, 3, 4, 5, 6, 7, 8
# m1 = 1
# m2 = 2
# o1 = 3
# o2 = 4
# s = [5, 6]
# m3 = 7
# m4 = 8
# key1 = key1
# key2 = key2
# b = 
# keys = {}

foo 1, 2, 3, 4, 5, 6, 7, 8, key1: 9
# m1 = 1
# m2 = 2
# o1 = 3
# o2 = 4
# s = [5, 6]
# m3 = 7
# m4 = 8
# key1 = 9
# key2 = key2
# b = 
# keys = {}

foo 1, 2, 3, 4, 5, 6, 7, 8, key1: 9, key2: 10
# m1 = 1
# m2 = 2
# o1 = 3
# o2 = 4
# s = [5, 6]
# m3 = 7
# m4 = 8
# key1 = 9
# key2 = 10
# b = 
# keys = {}

foo 1, 2, 3, 4, 5, 6, 7, 8, key1: 9, key2: 10, key3: 11
# m1 = 1
# m2 = 2
# o1 = 3
# o2 = 4
# s = [5, 6]
# m3 = 7
# m4 = 8
# key1 = 9
# key2 = 10
# b = 
# keys = {:key3=>11}

foo 1, 2, 3, 4, 5, 6, 7, 8, key1: 9, key2: 10, key3: 11, key4: 12 do end
# m1 = 1
# m2 = 2
# o1 = 3
# o2 = 4
# s = [5, 6]
# m3 = 7
# m4 = 8
# key1 = 9
# key2 = 10
# b = #<Proc:0x007fdc75135a48@(pry):77>
# keys = {:key3=>11, key4=>12}