Discussion on: 5 JavaScript "tips" that might bite you back.

[Josh]

I want union of any number of arrays, not just two. Any ideas how to refactor it still using the spread operator?

function union(a, ...b) {
  b.reduce((b, a) => [...a, ...b], a)
}

Granted, the spread operator being used in the returned array like this means you'll be generating copies upon copies upon copies, but you asked for an arity-agnostic refactor, not a performant one 😜

[EECOLOR]

Note that your ordering is a bit weird. You could do this (swap b and a in the reduce):

function union(...arrays) {
  return arrays.reduce((result, x) => [...result, ...x], [])
}

[Josh]

Every single time I use reduce, I forget how whichever language I'm using it in organizes either its own arguments, or the arguments given to its lambda/block. 🤦‍♂️

Ruby:

init = 0
(1..10).reduce(init) do |accumulator, iteration| 
  accumulator + iteration  
end # 55

Elixir:

init = 0
Enum.reduce(1..10, init, fn iteration, accumulator ->
  accumulator + iteration
end) # 55

And now, (thank you,) Javascript:

function union(a, ...b) {
  return b.reduce((accumulator, iteration) => [...accumulator, ...iteration], a)
}

[EECOLOR]

Haha, yeah. In Scala you have foldLeft and foldRight, where foldLeft has it as the first argument and foldRight as the second argument.

[Paul Smith]

This instantiates n+1 new arrays (where n is the number of arrays), when you could have just used a for loop to instantiate 1 new array.

[EECOLOR]

Did you read the comment of @josh to which I replied?

Granted, the spread operator being used in the returned array like this means you'll be generating copies upon copies upon copies, but you asked for an arity-agnostic refactor, not a performant one 😜