How to add all the arguments, wether there are two or more

[rianamoi]

Tell us what’s happening:
Hello,
I would like to add arguments together, if there is two arguments then add a+b, but if there is more than 2 arguments, the result here is undefined I don’t understand why ?
Could anyone please help. Thank you.
Your code so far

function addTogether() {
// Function to check if a number is actually a number
function checkNum(num) {
  return (typeof num === "number");
};

if (arguments.length > 1) {
  // Check if we have two arguments and if they are numbers
  // Return the sum if they are both numbers
  let args = [... arguments]
  if (checkNum(args)) {
    return args.reduce((a,b)=> a+b,0);
  } else {
    return undefined;
  }
} else if (arguments.length === 1) {
  // If only one argument was found, return a new function
  let first = arguments[0];
  if (checkNum(first)) {
    // Return function that expect a second argument.
    function addSecond(second) {
      // Check if the new argument is a number
      if (checkNum(second)) {
        return first + second;
      } else {
        return undefined;
      }
    };
    return addSecond;
  } else {
    return undefined;
  }
} 
}
console.log(addTogether(2, 3));

  **Your browser information:**

User Agent is: Mozilla/5.0 (Windows NT 6.1; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/89.0.4389.114 Safari/537.36.

Challenge: Arguments Optional

Link to the challenge:

freeCodeCamp.org

[Giles]

Hi there

Because checkNum takes a number, not an array. Anyway, since you’re already using the spread syntax you could simply do the following:

const addTogether = (...args) => (
  args
    .filter(item => typeof item === 'number')
    .reduce((accumulator, value) => accumulator + value, 0)
);

[ilenia]

@rianamoi

let args = [... arguments]
  if (checkNum(args)) {

this will never work, args is an array, so checkNum says false

[rianamoi]

Thank You for your time and commitment.