isNaN Check for Random Number Challenge, Part II

Question

For the Random Number Challenge, Part II video, I came up with a bit different solution that seems to work. 
In his solution, he uses:
javascript
if ( isNaN( lower ) || isNaN( upper ) ) {
}

my solution was to use:
javascript
if ( isNaN( lower || upper ) ) {
}

Is there a syntax or best-practice reason to rewrite the check twice?  We're really just testing a bunch of different inputs to make sure they are numbers. I find my way reads clearer to me, and would be less writing with more parameters. I just wanted to make sure there wasn't anything wrong with my method.

Taylor Michel · Accepted Answer

There actually is a difference in the two.  In your way you are using a shorthand statement where it will only return true if both values are NaN...  The reason for this being that inside of your isNaN call you have 'lower || upper'.  This is basically saying if 'lower' is NaN, null, or undefined use 'upper'.  Thus if lower is NaN then it will pass 'upper' into isNaN() and if lower isn't NaN it will only pass 'lower' into isNaN().
i.e.
```javascript
isNaN(24 || 24);
// isNaN = false ---> This will pass in the first 24 since it is valid, ignoring the second 24
isNaN(24 || NaN);
// isNaN = false ---> This will also pass in 24 since it is valid, ignoring the NaN
isNaN(NaN || 24);
// isNaN = false ---> This will see that NaN isn't valid and fallback on 24 passing that into isNaN, still resulting in false
isNaN(NaN || NaN);
// isNaN = true ---> This is the only case in which it will result in true because the first NaN fails and it falls back on the second one passing NaN into the isNaN function.
```
Hopefully this helps in understanding why the first way is the correct way of checking.
Taylor

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isNaN Check for Random Number Challenge, Part II

1 Answer

Taylor Michel

Taylor Michel

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