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Start your free trialDan MacDonagh
4,615 PointsReturning variable instead of returning calculation
In the beginning part of the video, when he changed the function to return a value, he assigned the Math.floor() calculation into a variable, and then returned that variable. Why couldn't you just return the Math.floor calculation inside the function and call it that way?
I tested it out, and this:
function getRandomNumber() {
return Math.floor( Math.random() * 6 ) + 1;
}
alert(getRandomNumber());
acted in the same way as this:
function getRandomNumber() {
var randomNumber Math.floor( Math.random() * 6 ) + 1;
return randomNumber;
}
alert(getRandomNumber());
Is there a difference in the methods I'm not seeing?
1 Answer
Tony Nguyen
24,934 PointsYou're thinking more like a programmer now! Yes, those both work and provide the same results and there isn't a difference in terms if functionality as they both do the same thing. I think the instructor was just doing that to provide more readability and understanding.
Dan MacDonagh
4,615 PointsDan MacDonagh
4,615 PointsThanks for the quick response! Glad I wasn't going crazy :)