Use recursion to create countdown help

JavaScript
#1

Hello Guys ,
this is my code :

const myArray = [];
function countdown(myArray, n) {
  if (n <= 0) {
    return;
  } else {
    myArray.push(n);
    countdown(myArray, n - 1);
  }
  return myArray;
}
countdown(myArray, 5);

and i don’t know why it’s not passing , while when i wrote it in the visual code it worked perfectly , any idea why it’s not working ?

Basic JavaScript: Use Recursion to Create a Countdown
#2

In the future, please create your own topic when you have specific questions about your own challenge code. Only respond to another thread when you want to provide help to the original poster of the other thread or have follow up questions concerning other replies given to the original poster.

The easiest way to create a topic for help with your own solution is to click the Ask for Help button located on each challenge. This will automatically import your code in a readable format and pull in the challenge url while still allowing you to ask any question about the challenge or your code.

Thank you.

#3

I’ve edited your post for readability. When you enter a code block into a forum post, please precede it with a separate line of three backticks and follow it with a separate line of three backticks to make it easier to read.

See this post to find the backtick on your keyboard. The “preformatted text” tool in the editor (</>) will also add backticks around text.

Note: Backticks are not single quotes.

markdown_Forums

#4

Your code contains global variables that are changed each time the function is run. This means that after each test completes, subsequent tests start with the previous value. To fix this, make sure your function doesn’t change any global variables, and declare/assign variables within the function if they need to be changed.

Example:

var myGlobal = [1];
function returnGlobal(arg) {
  myGlobal.push(arg);
  return myGlobal;
} // unreliable - array gets longer each time the function is run

function returnLocal(arg) {
  var myLocal = [1];
  myLocal.push(arg);
  return myLocal;
} // reliable - always returns an array of length 2
1 Like
#5

i didn’t notice that , well thank you so much for the explanations , saves more time , also thank you for the help

#6