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JavaScript

SeHyun Choi
SeHyun Choi
3,441 Points
Posted by SeHyun Choi
SeHyun Choi
3,441 Points

Return in Javascript function

I have question about following code...

function getRandomNumber( upper ) { var randomNumber = Math.floor( Math.random() * upper ) + 1 ; return randomNumber;

}

console.log(getRandomNumber(54));

so basically what "return" is doing is....

It exits the function called getRandomNumber and then get the parameter in this case 54 and then bring it back to variable randomNumber?

Am I understanding this correctly?

1 Answer

Steven Parker
Steven Parker
229,644 Points
Steven Parker
Steven Parker
229,644 Points

The "return" keyword does two things. It ends the function, but it also passes back a value to the point in the code where the function was called.

The value passed back in this case is whatever is stored in the variable "randomNumber". We can't predict what the value will be even if we know what "upper" is because of the "random" part of the formula.

SeHyun Choi
SeHyun Choi
3,441 Points
SeHyun Choi
SeHyun Choi
3,441 Points

Thank you for the reply. But I am still confused..

function getRandomNumber( upper ) { var randomNumber = Math.floor( Math.random() * upper ) +1;

}

alert(getRandomNumber(6));

when I do this why do I get alert message saying "undefined"?

I am confused because in original question I typed return randomNumber. Then for console.log() I typed getRandomNumber instead of randomNumber. Please help I'm so confused.

Steven Parker
Steven Parker
229,644 Points
Steven Parker
Steven Parker
229,644 Points

This version has no "return". Remember that the second thing a "return" does is to pass back the value. So when no value is passed back to the "alert", it shows "undefined".