Question 374636 in Javascript Question posted in
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Javascript – Why does an array of even numbers contain empty values?

Why does an array of even numbers contain empty values?

function getEvenNumbers(arr) {
  const evenNumbers = [];
  for (let i = 0; i < arr.length; i++) {
    if (arr[i] % 2 === 0) {
      evenNumbers[i] = arr[i]
    }
  }
  return evenNumbers;
}
console.log(getEvenNumbers([1, 2, 3, 4, 5, 6]));

console output: [empty, 2, empty, 4, empty, 6]

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2

Answers


  1. 0

    it’s because you’re assigning at index i

    function getEvenNumbers(arr) {
    const evenNumbers = [];
    for (let i = 0; i < arr.length; i++) {
        if (arr[i] % 2 === 0) {
            evenNumbers[i] = arr[i]
        }
    }
    return evenNumbers;
}
console.log(getEvenNumbers([1,2,3,4,5,6]));

    or given this array

    evenNumbers[1] = 2
evenNumbers[3] = 4
evenNumbers[5] = 6

    with javascript filling up the empty space. you should use evenNumbers.push(arr[i]) instead

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  2. 0

    It is because you are assigning a value to the i-th index of the array.
    Since the odd value is skipped, they appear to be empty.
    You can push even numbers to the evenNumbers array.

    function getEvenNumbers(arr) {
    const evenNumbers = [];
    for (let i = 0; i < arr.length; i++) {
        if (arr[i] % 2 === 0) {
            evenNumbers.push(arr[i])
        }
    }
    return evenNumbers;
}
console.log(getEvenNumbers([1,2,3,4,5,6]));

    Even more, you can use Array.filter() function to do the same work.

    function getEvenNumbers(arr) {
    return arr.filter(n=>n%2===0);
}
console.log(getEvenNumbers([1,2,3,4,5,6]));
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