JavaScript function - named parameters issue when skipping an optional parameter

Question

If you run the below code in browser's console:

function named_param(a, b=null, c=5) {
  console.log("a="+a);
  console.log("b="+b);
  console.log("c="+c);
}

named_param(3, c=10)

The output received is:

a=3
b=10
c=5

The output I am looking for, is:

a=3
b=null
c=10

I saw the below two URLs and tried a few workaround but it didn't work

Javascript Function Parameters Not Working

Named parameters in javascript

Let me know the steps I am doing wrong here. Tried the below codes but didn't work: A:

function named_param(a, {b=null, c=5}) {
  console.log("a="+a);
  console.log("b="+b);
  console.log("c="+c);
}

named_param(3, c=10)

B:

function named_param(a, {b=null, c=5}={}) {
  console.log("a="+a);
  console.log("b="+b);
  console.log("c="+c);
}

named_param(3, c=10)

JavaScript does not have named parameters. Please try running `named_param(3, c=10)` in strict mode and you'll get an exception. — Bergi, Jun 30 '19 at 18:24
The "*Javascript Function Parameters Not Working*" question has nothing to do with this. — Bergi, Jun 30 '19 at 18:27

score 1 · Answer 1 · answered Jun 30 '19 at 18:28

You can skip parameter in between, so when you pass

named_param(3, c=10)
            |   |_____________ considered as `b`
            |_________________ considered as `a`

You can use object and destructuring

function named({a,b=null,c=3} = {}){
  console.log('a-->',a)
  console.log('b-->',b)
  console.log('c-->',c)
}

named({a:1,c:2})
named()
named({c:2})

score 1 · Accepted Answer · answered Jun 30 '19 at 18:29

When using the destructuring solution, you must pass an actual object. named_param(3, c=10) simply is invalid syntax (well, it's valid, but it's an assignment to a global c variable that is passed as the second argument - no naming).

function named_param(a, {b=null, c=5}={}) {
  console.log("a="+a);
  console.log("b="+b);
  console.log("c="+c);
}

named_param(3)
named_param(3, {b:"hi"})
named_param(3, {c:10})
named_param(3, {b:"hi", c:10})

score 1 · Answer 3 · answered Jun 30 '19 at 18:31

1

What you can do is something using destructuring objects like:

function named_params({a = 4, b = null, c = 5} = {a:1, b: 2, c:3}) { console.log(a,b,c) }

then, the call for your function will be:

named_params({a: 3, c:10})

answered Jun 30 '19 at 18:31

Daniel Peña

76
5

Having different default values for `name_params()` vs `name_params({})` is weird – Bergi Jun 30 '19 at 18:40

Alexander Zinchenko · Answer 4 · 2019-06-30T18:55:31.900

Well, it's not an issue, it's just you trying to use named parameters where they are not supported at all.

To reach what you need, you should do like this:

function named_param({a, b, c}) {
  console.log("a="+a);
  console.log("b="+b);
  console.log("c="+c);
}

named_param({a:5, c:9}); 
// Output:
// a=5
// b=undefined
// c=9

At first I misunderstood your question and proposed to use destructuring assignment with arguments object.

JavaScript function - named parameters issue when skipping an optional parameter

4 Answers4