How to pass the nth optional argument without passing prior arguments in JavaScript functions?

Question

So imagine we got a function like below, and I want to use the default value for the first parameter a and a custom value for second parameter b.

const func = (a = 1 , b = 2) => {
    console.log("a is equal to " + a, " and b is equal to " + b)
}

I want to just pass one value for parameter b. something like this:

func(b : 7);
// I want it to print: a is equal to 1 and b is equal to 7

How can I achieve it in JavaScript?

If you don't want to change the function signature, you would need to call it like so `func(undefined, 7)` — Jake, Jul 15 '19 at 13:17

Nenad Vracar · Accepted Answer · 2019-07-15T13:16:02.837

0

You modify your function to receive an object and then inside that object you can also define default values for some properties like a and b.

const func = ({ a = 1, b = 2 }) => {
  console.log("a is equal to " + a, " and b is equal to " + b)
}

func({b: 3})

edited Jul 15 '19 at 13:16

answered Jul 15 '19 at 13:13

Nenad Vracar

So how about methods like `server.listen(port, hostname, backlog, callback)`? all its parameters are optional. – mgh Jul 15 '19 at 13:15
Unless the function can receive an object where you can specify name of the parameter you want to pass there is no way to use named parameters. – Nenad Vracar Jul 15 '19 at 13:17

score 0 · Answer 2 · answered Jul 15 '19 at 13:14

0

try to use RORO pattern, to pass object into function

answered Jul 15 '19 at 13:14

Pasha K

2 Answers2