Hash Destructuring

Question

You can destructure an array by using the splat operator.

def foo(arg1, arg2, arg3)
  #...Do Stuff...
end
array = ['arg2', 'arg3']
foo('arg1', *array)

But is there a way to destruct a hash for option type goodness?

def foo(arg1, opts)
  #...Do Stuff with an opts hash...
end
opts = {hash2: 'bar', hash3: 'baz'}
foo('arg1', hash1: 'foo', *opts)

If not native ruby, has Rails added something like this?

Currently I'm doing roughly this with

foo('arg1', opts.merge(hash1: 'foo'))

Busy

If you are talking about default option, yes merge is the way to go. – Unspent

Is there any reason that you reversed the order instead of doing {hash1: 'foo'}.merge(opts)? – Jehu

@Jehu Not really. Just what came out naturally. – Busy

Answer 1

It's 2018 and this deserves an update. Ruby 2.0 introduced keyword arguments and with that also the hash splat operator **. Now you can simply do the following:

def foo(arg1, opts)
  [arg1, opts]
end

opts = {hash2: 'bar', hash3: 'baz'}
foo('arg1', hash1: 'foo', **opts)
#=> ["arg1", {:hash1=>"foo", :hash2=>"bar", :hash3=>"baz"}]

Twoup

Keep in mind that this only works with symbol keys. {**{a: 1}, **{b: 2}} #=> {:a=>1, :b=>2} while {**{'a' => 1}, **{'b' => 2}} #=> TypeError (wrong argument type String (expected Symbol)) – Twoup

Yep. I was looking to add some keys to a model.as_json call and used Rails' symbolize_keys method: hash = {a: 1, b: 2, **model.as_json.symbolize_keys} – Gemmell

Answer 2

Yes, there is a way to de-structure a hash:

def f *args; args; end
opts = {hash2: 'bar', hash3: 'baz'}
f *opts  #=> [[:hash2, "bar"], [:hash3, "baz"]]

The problem is that you what you want is actually not de-structuring at all. You’re trying to go from

'arg1', { hash2: 'bar', hash3: 'baz' }, { hash1: 'foo' }

(remember that 'arg1', foo: 'bar' is just shorthand for 'arg1', { foo: 'bar' }) to

'arg1', { hash1: 'foo', hash2: 'bar', hash3: 'baz' }

which is, by definition, merging (note how the surrounding structure—the hash—is still there). Whereas de-structuring goes from

'arg1', [1, 2, 3]

to

'arg1', 1, 2, 3

Orland

Answer 3

There is no such thing (although it has been proposed). Since this will change the parsing rule, it cannot be implemented within Ruby. The best I can think of is to define * on hash like

class Hash; alias :* :merge end

and use it in one of the following ways:

foo('arg1', {hash1: 'foo'}*opts)
foo('arg1', {hash1: 'foo'} *opts)
foo('arg1', {hash1: 'foo'}. *opts)

the last of which I think is reasonably close to what you wanted.

Jehu

Perhaps using + instead of * would make more sense? – Orland

@AndrewMarshall I agree with that. I wanted to make it look close to the splat operator. – Jehu

It's now possible in Ruby 3, though with a different/new syntax: {a: 1, b: 2, c: 3, d: 4} => {a:, b:, **rest} (see ruby3.dev/ruby-3-fundamentals/2021/01/06/…) – Scolopendrid

Answer 4

If you're ok with using active_support:

require 'active_support/core_ext/hash/slice.rb'

def foo(*args)
  puts "ARGS: #{args}"
end

opts = {hash2: 'bar', hash3: 'baz'}
foo *opts.slice(:hash2, :hash3).values

...or you could monkey-patch your own solution:

class Hash
  def pluck(*keys)
    keys.map {|k| self[k] }
  end
end

def foo(*args)
  puts "ARGS: #{args}"
end

opts = {hash2: 'bar', hash3: 'baz'}
foo *opts.pluck(:hash2, :hash3)

Chile