Help woth FindersKeeper

JavaScript
#1

I was wondering why i get an error here - i know that by omitting === true and else I get the answer but I’d like to understand the logic behind that.

Thanks from Belgium

findElement([1, 3, 5, 8, 9, 10], function(num) { return num % 2 === 0; }) should return 8.
Passed :white_check_mark:

findElement([1, 3, 5, 9], function(num) { return num % 2 === 0; }) should return undefined. :x:

function findElement(arr, func) {
  let num = 0;

  for (var i = 0; i < arr.length; i++) {
    num = arr[i];
    if (func(num) === true) {
      return num;
    }
    else {
    return undefined;
}
  }

  
}
Basic Algorithm Scripting: Finders Keepers
#2

so, what happens if for arr[0] the function output false?

#3

function findElement(arr, func) {

let num;

for ( let no of arr){

  if (func(no)){

    return no;

  }

}

return undefined;

}

findElement([1, 2, 3, 4], num => num % 2 === 0);

#4 
if (func(num) === true) {
      return num;
 } else {
    return undefined;
 }

Your code reads as following: If the function returns true return my number, else return undefined. The problem is that if the first comparison func(num) === true fails, the else branch is executed and your code returns undefined, so you don’t traverse the whole array, you stop at the first element.

The correct code below:

You should move the return from the else branch outside the for loop, so if you don’t find a case that passes in the whole array, return undefined

function findElement(arr, func) {
  let num;

  for (var i = 0; i < arr.length; i++) {
    num = arr[i];
    if (func(num) === true) {
      return num;
    }
  }

  return undefined;
}
1 Like
#5

Thanks so much for this clear reasoning :blush:

#6