Set only one optional parameter of a JS function with ES6

Question

In python, we can do:

func(a, b=1, c=2)

And then call it like that

func(8, c=5)

With ES6, we can declare it in the same way :

func(a, b=1, c=2)

But how to set c without setting b to undefined ?

Answer 1

You can create a function that accepts an object, and get the parameters using destructuring. Now you can pass the parameters in any order you wish:

function fn({ a, b = 1, c = 2 }) {
  console.log({ a, b, c })
}

fn({ a: 8, c: 5 })