Is this method fine?

Question

function getRandomNumber( lower, upper ) { if ( isNan = true) { throw new Error('NOT A NUMBER'); }

I did this and got the error that I needed to get, but he mentioned needing to use ||, which I didnt do, nor could I figure out how

Answer 1 · 2019-03-02T18:06:18Z

March 2, 2019 6:06pm

Unfortunately, no. Copying your function into multiple lines to see it better:

function getRandomNumber( lower, upper ) {
  if ( isNan = true) {
    throw new Error('NOT A NUMBER');
  }

Several things here:

You haven't closed your function with a curly brace. Hopefully this is just a copying omission.
You're not using either argument (lower and upper) you pass into the function. If you don't need them, don't pass them in.
This function will always throw a NOT A NUMBER error, because your if statement isn't evaluating anything, it's assigning something. To evaluate something, you have to use double- or triple-equals: a == b or a === b. This compares a and b to each other. If you say a = b, you are assigning the value of b as the value of a, the same way you'd say a = 3 to set the variable's value. Assignments of positive values always return true, so your if statement will always run.
isNaN first off needs a capital N at the end, and also is not a standalone value, it's a function. You have to call it and you have to pass a value to it: isNaN(3) or isNaN(a). If you do that, it will return true or false. Extra note: since the isNaN() function itself returns true or false, you do not have to compare it to true at all in the if function. You'd usually see this written like so: if ( isNaN(a) ) { /* whatever */ } since isNaN returning true or false will give the if condition all it needs to evaluate whether to run or not.

Hope that all makes sense! Cheers.

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