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Cuéntanos qué está pasando:
Describe tu problema en detalle aquí.
No estoy entendiendo para modificar la funcion con parametros rest y estuve buscando informacion y no logro entenderlo…!!! si alguien me podria ayudar.

  **Tu código hasta el momento**

const sum = (x, y, z) => {
const args = [x, y, z];
return args.reduce((a, b) => a + b, 0);
}
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Desafío: Utiliza el parámetro rest con parámetros de función

Enlaza al desafío:

The rest operator will gather the remaining items into an array. So, if I have a function like:

const concatenateThreeStrings = (str1, str2, str3) => {
  return str1 + str2 + str3;
}

That works, but what about using the rest operator instead:

const concatenateThreeStrings = (...strArr) => {
  return strArr[0] + strArr[1]  + strArr[2] ;
}

Actually, now that they are in an array, I can just use join:

const concatenateStrings = (...strArr) => {
  return strArr.join('');
}

Or even simplify to:

const concatenateStrings = (...strArr) => strArr.join('');

Not only is that simpler, but notice that it will work with any number of strings.


You want to do something similar. You have:

const sum = (x, y, z) => {
  const args = [x, y, z];
  // ...

You are getting in each argument and then manually putting them in an array. With the rest operator, you can do both in one step, like I did in the examples above. That is what the rest operator does - it takes a list of elements and puts them into an array.

Does that help?