Report even or odd number in JS

Question

I'm trying to do the following exercise: Create a function that takes an integer as an argument and returns "Even" for even numbers or "Odd" for odd numbers.

Here is my code; I can't figure out what I'm missing:

function even_or_odd(n) {
if(n % 2 === 0)
{
  console.log('Even');
}
else
{
  console.log('Odd');
}
};

Answer 1

You are not returning anything, just outputing it in the console.

function even_or_odd(n) {
    if(n % 2 === 0) {
      return 'Even';
    } else {
      return 'Odd';
    }
};

Answer 2

假设“i”是要检查的数字...

if(i%2){alert('odd');} else {alert('even');}

Answer 3

Nothing just return the string if returning is the goal:

 function even_or_odd(n) { if (n % 2 === 0) { return 'Even'; } else { return 'Odd'; } }; alert(even_or_odd(5)); alert(even_or_odd(2));

Answer 4

If anybody looking for shorthand using es6, They can try this-

const evenOrOdd = number => number % 2 ? 'Odd' : 'Even';

// Outputs-
evenOrOdd(2) // Even
evenOrOdd(3) // Odd

Answer 5

yeha I was having the same problem,then i remembered i need to return the values

function even_or_odd(number) {
      if(number % 2 == 0){
        return ("Even");
      }else{
        return ("Odd");
      }
}