How create new instance of a Ruby Struct using named arguments (instead of assuming correct order of arguments)

Question

Given: Customer = Struct.new(:name, :address, :zip)

Is there a way to name the arguments instead of presuming a certain order?

The docs say do it like this:

joe = Customer.new("Joe Smith", "123 Maple, Anytown NC", "12345")

which IMO makes it too easy to switch two parameters accidentally.

I'd like to do something like this:

joe = Customer.new(name: "Joe Smith", address: "123 Maple, Anytown NC", zip: "12345")

so that the order is not important:

joe = Customer.new(zip: "12345", name: "Joe Smith", address: "123 Maple, Anytown NC")

If it doesn't have to be a struct you could consider ostruct, e.g., stackoverflow.com/a/11962369/438992. There are performance issues, but they don't necessarily matter. — Dave Newton
– Dave Newton, Commented Jun 1, 2015 at 17:23
i considered openstruct but AFAIK it does not provide the protection against accidentally mis-named keys, for example one could accidentally set namezzzz: "John doe" and it would create a new (wrong) attribute, whereas if you tried that with a Struct it would alert you by throwing an exception (as desired). — jpw
– jpw, Commented Jun 1, 2015 at 20:13

pgblu · Accepted Answer · 2015-06-01 17:28:48Z

2

Named parameters are not (yet) possible in Ruby's Struct class. You can create a subclass of your own in line with this Gist: https://gist.github.com/mjohnsullivan/951668

As you know, full-fledged Classes can have named parameters. I'd love to learn why they aren't possible with Structs... I surmise that someone on the Core team has thought of this and rejected it.

answered Jun 1, 2015 at 17:28

pgblu

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2 Comments

engineersmnky Over a year ago

The NamedStruct is nice although I think *args is not needed since you are excluding everything after the first argument. eg. def initialize attributes would function equivalently without having to pack them into an Array and to_hash could be simplified to members.each_with_object({}) {|m,obj| obj[m] = self[m]}

pgblu Over a year ago

Good point, yes. To be clear, it's not my gist - credit where credit's due.

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