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Seokhyun Wie
Full Stack JavaScript Techdegree Graduate 21,606 PointsUsing 'else' clause instead of directly writing 'return' clause
Hi guys, I used a little bit of different approach from the solution that is mentioned in the video.
function randomNumber(lower, upper) {
if (isNaN(lower) || isNaN(upper)) {
throw new Error('error message');
} else {
return Math.floor(Math.random() * (upper - lower + 1) + lower);
}
}
console.log(randomNumber('nine', 1000));
Is there any problem with this approach? If there is, please let me know, or if not, also please let me know the pros and cons of it. Thank you so much!
2 Answers
Dmitry Polyakov
4,989 PointsThese two functions will produce the same result
function randomNumber(lower, upper) {
if (isNaN(lower) || isNaN(upper)) {
throw new Error('error message');
} else {
return Math.floor(Math.random() * (upper - lower + 1) + lower);
}
}
console.log(randomNumber('nine', 1000));
function randomNumber(lower, upper) {
if (isNaN(lower) || isNaN(upper)) {
throw new Error('error message');
}
return Math.floor(Math.random() * (upper - lower + 1) + lower);
}
console.log(randomNumber('nine', 1000));
Steven Parker
229,644 PointsAdding the "else" won't cause any difference in behavior; but it's not needed, since when the "if" test is true, the "throw" will stop the function. So any code after the conditional block can only run in the "else" condition.
Seokhyun Wie
Full Stack JavaScript Techdegree Graduate 21,606 PointsSeokhyun Wie
Full Stack JavaScript Techdegree Graduate 21,606 PointsThansk Dmistry, I just checked it works the same. However I wonder what would be the mere difference there. :)