ES6: Use the Rest Operator with Function Parameters stuck

JavaScript
1

Tell us what’s happening:
I keep getting caught on the very last objective where I am to use the ... rest parameter on the args parameter. What do I need to do to pass that test?

Your code so far


const sum=function sum(...args){
return args.reduce((a,b)=>a+b, 0);
}
console.log(sum(1, 2, 3)); // 6

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Challenge: Use the Rest Parameter with Function Parameters

Link to the challenge:
https://www.freecodecamp.org/learn/javascript-algorithms-and-data-structures/es6/use-the-rest-parameter-with-function-parameters

1 Like
3

I’ve tried adding the rest parameter to the original provided code by adding ... to x, y, z when the function is first being described. It does not work. Could you point me in the correct direction in more detail?

5

Please don’t laugh, but my first thought was very simplistic and that was just to add ... to the first variable described as shown below.

const sum = (…x, y, z) => {
const args = [x, y, z];
return args.reduce((a, b) => a + b, 0);
}
console.log(sum(1, 2, 3)); // 6

I’ve also attempted adding ... before each of the x, y and z when sum is first defined and also when args is first defined. The code that I ultimately shared at the start of my post was the only one that could pass all the tests except the last one.

7

The trouble is, the example only lists one parameter in the initial definition of the function ...args. I don’t really know how to translate that onto the problem presented. Am I supposed to leave (x, y, z) alone in the solution?

I have also tried (x, y, …z) and that didn’t seem to work either.

1 Like
10

Then why would (…arg) be wrong here?

12

but arg isn’t a variable. It’s a parameter isn’t it?
In the example, ...arg isn’t defined as a variable either. So it’s quite confusing.

14

Still confused.

function howMany(...args) {
  return "You have passed " + args.length + " arguments.";
}
console.log(howMany(0, 1, 2)); // You have passed 3 arguments.
console.log(howMany("string", null, [1, 2, 3], { })); // You have passed 4 arguments.

args is never defined here.

16

What’s the difference between the example code and mine?

I don’t see any const args= or var args= in the example code, but I think you’re implying that that part needs to be in mine despite it.

18

I don’t understand your original correction then. If I don’t need to separately declare args as a variable, then what is the problem with the original code?

20

I don’t understand your first reply.

22

so the very first line is what I understand to be the function parameters.

const sum = (x, y, z) =>

I am to only change the x, y and z of this line?

24

It’s the format that’s shown in the “Get a hint” link. It’s the firs ttime where the solution shown is not the one that works.

25

Okay, finally got it.

const sum = (…args) => {

return args.reduce((a, b) => a + b, 0);

}

console.log(sum(1, 2, 3)); // 6

it works on the site. Is this what you were expecting as well?

1 Like
28

Okay, thank you very much for your help!

30

Thank you very much for this!