How to return first odd number - JavaScript-CodePudding

I can't figure out how to make my code return the first odd number and if there are no odd numbers to return undefined. What am I doing wrong/need to research? The 3 requirements I have are;

1 Return the first odd number in an array of odd numbers

2 Return the first odd number when it is the last number in the array

3 Return undefined when there are no odd numbers

function firstOdd(arg){
  for (let i=0; i <= arg.length; i  ) 
  if (i % 2 !== 0) {
    return(i)
  } else {
    return undefined;
  }
}

Thank you in advanced for being kind with me.

CodePudding user response：

Take return undefined out of the loop. It's only true if none of the numbers are even. Then, make sure you test the values in the array with arg[i]. You were using the index i itself.

function firstOdd(arg){
  for (let i=0; i < arg.length; i  ) {
    if (arg[i] % 2 === 1) {
      return arg[i]
    }
  }
  return undefined;
}

console.log(firstOdd([1,3,5,7,9]))
console.log(firstOdd([2,4,6,8,9]))
console.log(firstOdd([2,4,6,8,10]))

CodePudding user response：

You can use the following code . As array starts from index 0 , so the last index of the array will be (arg.length-1).That's why I am iterating till the index is less than array length. As soon it finds odd numbers it return that and function terminates . If there is no odd number , by default the function will return 'undefined' . Hope you will understand .

function firstOdd(arg) {
  for (let i = 0; i < arg.length; i  )
    if (i % 2 !== 0) {
      return i;
    } else {
      continue;
    }
  return undefined;
}