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Exercise 3

Let's improve our previous implementation of evenOrOdd. Add a validation check to ensure that the argument is an integer. If it isn't, the function should issue an error message and return.

You can use Number.isInteger(value) to determine whether a numeric value is an integer. It returns true if the value is an integer, false otherwise.

Solution

function evenOrOdd(number) {
  if (!Number.isInteger(number)) {
    console.log('Sorry, the value you passed is not an integer');
    return;
  }

  if (number % 2 === 0) {
    console.log('even');
  } else {
    console.log('odd');
  }
}

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