Welcome to the Treehouse Community
Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.
Looking to learn something new?
Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.
Start your free trialHina K.
4,806 PointsMy solution (different from the tutorial). Does this work? I used the '&&' operator.
const random = (lowNum, highNum) => {
if (!isNaN(lowNum) && !isNaN(highNum)) {
return Math.floor((Math.random() * (highNum - lowNum + 1) + lowNum));
} else {
console.log("Enter two numbers and try again.");
}
}
** Call functions I used **
console.log(random(10, 30));
console.log(random(20, 40));
console.log(random('a', 40));
console.log(random(0, 2));
1 Answer
Steven Parker
231,122 PointsInverting the logic to test for a good condition instead of a bad one is perfectly legitimate from a functional perspective, and a good example of how there's rarely just one solution when programming. Good job!