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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
231,236 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!