Default descontructed parameter on function

Question

I am trying to use a function that works for two scenarios, it can either iterate through the key of an object in array or the array value itself. So this prints fine:

const arr1 = [{
  car: 'audi'
},{
  car: 'bmw'
}]

const printCars = arr1.map(({ car }) => console.log(car))
// audi
// bmw

But I dont know how to use the same function's default parameter to allow for a simple array

const arr2 = ['audi', 'bmw'];

const printCars = arr2.map(({ car }) => console.log(car));
// Obviously car is undefined

I wonder how can I do something like

const arr = ['audi', 'bmw'];
const printCars = arr.map(({ car } = car) => console.log(car));
// audi
// bmw

Answer 1

I don't think there's a good way to do this in just the parameter list. You can destructure on the first line of the function, and assign a default value of the whole item:

const arr2 = ['audi', 'bmw'];

arr2.forEach((item) => {
  const { car = item } = item;
  console.log(car);
});

Note that since you aren't creating a new array, but performing side-effects (console.log), you should not use .map - instead, use forEach. .map is only for when you need to transform every element of one array into a new element in a different returned array.

It would be possible to put the proper value into the car variable using just the parameter list, but it's quite ugly and an abuse of default arguments, I'll put it below for informational purposes only (please don't use it):

const arr2 = ['audi', 'bmw'];

arr2.forEach((item, i, arr, { car = item } = item) => console.log(car));

The above works because forEach only takes 3 arguments, so the 4th will be undefined, so you can use default arguments to put something into the car identifier conditionally.