Question 176716 in Javascript Question posted in
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Why does my Javascript function return undefined?

I am trying to write a function that returns the first odd number in an array.

I have this so far but it is returning undefined.

function findFirstOdd(numbers) {
  let oddNumbers = [];
  for (let i = 0; i <= numbers.length; i++) {
    if (numbers[i] % 2 != 0) {
      oddNumbers.push(numbers[i])
    } else {}
  };
  return oddNumbers[0]
};

console.log(findFirstOdd(6, 2, 9, 3, 4, 5));

I would expect this to return 9; being the first odd number.

Tags:

4

Answers


  1. 0

    Your function call is wrong, numbers should be in array.

    console.log (findFirstOdd([6, 2, 9, 3, 4, 5]));
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  2. 0

    Your code expects to receive an array as argument, but you are passing multiple numbers, not an array.
    There are two simple ways to solve this:

    function findFirstOdd(...numbers) {

    OR

    console.log(findFirstOdd([6, 2, 9, 3, 4, 5]));

    Also, one suggestion, you can remove empty else + you may return first odd number right after you found it
    Full code:

    function findFirstOdd(numbers) {
  for (let i = 0; i <= numbers.length; i++) {
    if (numbers[i] % 2 !== 0) {
      return numbers[i];
    }
  }
  return undefined;
}

console.log(findFirstOdd([6, 2, 9, 3, 4, 5]));

    UPD:
    After you learn array methods, you can upgrade it to:

    function findFirstOdd(numbers) {
  return numbers.find((number) => number % 2 !== 0);
}

console.log(findFirstOdd([6, 2, 9, 3, 4, 5]));
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  3. 0

    numbers parameter most be an array:
    I cleaned your script into this:

    function findFirstOdd(numbers) {
  let oddNumbers = [];
  for(let i = 0; i < numbers.length; i++) {
    if (numbers[i] % 2 !== 0){
      oddNumbers.push(numbers[i]);
    } else {
      //do nothing
    }
  }
  return oddNumbers;
};

var numbers = [6, 2, 9, 3, 4, 5];

console.log (findFirstOdd(numbers));

    I change some sections in your code:

    I changed entry of function into an array: var numbers = [6, 2, 9, 3, 4, 5];

    for(let i = 0; i <= numbers.length; i++) to for(let i = 0; i < numbers.length; i++)

    return oddNumbers[0] to return oddNumbers;

    semicolon ; on javascript codes are not necessary.

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  4. 0

    JavaScript functions arguments can only relate to a single variable.

    In your case you have 2 choices:

    Rest parameters usage:

    function findFirstOdd(...numbers) {
  let oddNumbers = [];
  for (let i = 0; i <= numbers.length; i++) {
    if (numbers[i] % 2 != 0) {
      oddNumbers.push(numbers[i])
    } else {}
  };
  return oddNumbers[0]
};

console.log(findFirstOdd(6, 2, 9, 3, 4, 5));

    keyword arguments usage:

    function findFirstOdd() {
  let oddNumbers = [];
  for (let i = 0; i <= arguments.length; i++) {
    if (arguments[i] % 2 != 0) {
      oddNumbers.push(arguments[i])
    } else {}
  };
  return oddNumbers[0]
};

console.log(findFirstOdd(6, 2, 9, 3, 4, 5));

    PS: one line code ( odd numbers right bit have a binary value === 1 )

    const findFirstOdd = (...n) => n.find( n => n & 1);  

console.log(findFirstOdd(6, 2, 9, 3, 4, 5));
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