I have written a function to search an array and log a number(x) if it is found. The code below works as expected and 1 is logged.

let myArr = [54, 23, 499, 342, 1, 44];
let x = 1;

let searchNumber = (arr) => {
  arr.forEach(ele => {
    if (ele == x) {
      console.log(ele);
    }
  })
};

searchNumber(myArr);

I'd like to now return the number that I assign to variable x. Expecting y to be assigned 499 but when I log y it returns undefined. Where am I going wrong?

let myArr = [54, 23, 499, 342, 1, 44];
let x = 499;

let searchNumber = (arr) => {
  arr.forEach(ele => {
    if (ele == x) {
      return ele;
    }
  })
};

let y = searchNumber(myArr);

I have written a function to search an array and log a number(x) if it is found. The code below works as expected and 1 is logged.

let myArr = [54, 23, 499, 342, 1, 44];
let x = 1;

let searchNumber = (arr) => {
  arr.forEach(ele => {
    if (ele == x) {
      console.log(ele);
    }
  })
};

searchNumber(myArr);

I'd like to now return the number that I assign to variable x. Expecting y to be assigned 499 but when I log y it returns undefined. Where am I going wrong?

let myArr = [54, 23, 499, 342, 1, 44];
let x = 499;

let searchNumber = (arr) => {
  arr.forEach(ele => {
    if (ele == x) {
      return ele;
    }
  })
};

let y = searchNumber(myArr);

• javascript

• you can use Array.prototype.find to return the first element that satisfies a condition. – Lucas D Commented Oct 31, 2020 at 5:51

5 Answers 5

return ele inside forEach callback is not the return of the searchNumber function.

forEach executes a provided function once for each array element so return ele inside that will act like return to that provided function inside forEach.

It does not repesent the return to the main function.

In this case, it's better to use for loop.

let myArr = [54, 23, 499, 342, 1, 44];
let x = 499;

let searchNumber = (arr) => {
  for (let i = 0; i < arr.length; i ++) {
    if (arr[i] == x) {
      return arr[i];
    }
  }
};

let y = searchNumber(myArr);
console.log(y);

If you look at the placement of the return statement, it is actually within the arrow function being called for each element (ele => {...}). No value is actually being returned from the scope of the function searchNumber.

Try creating a variable in the scope of searchNumber and modifying it from arr.forEach() instead:

let myArr = [54, 23, 499, 342, 1, 44];
let x = 499;

let searchNumber = (arr) => {
    let val = null;
    arr.forEach(ele => {
        if (ele == x) {
            val = ele;
        }
    });
   return val;
};

let y = searchNumber(myArr);

You cannot break (exit) a forEach, it will iterate all elements. the array.find() should do the trick https://www.w3schools./jsref/jsref_find.asp or array.findIndex() https://www.w3schools./jsref/jsref_findindex.asp

.find will return the matching element:

let myArr = [54, 23, 499, 342, 1, 44];
let x = 499;

let searchNumber = (arr) => arr.find(ele => ele == x);

let y = searchNumber(myArr);

console.log(y);

what you are looking here is probably the find function.

let searchNumber = (arr) => arr.find(e => e === x);

let y = searchNumber(myArr);