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'))
Answer
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