Javascript: confusion on using Logical NOT ! Operator [duplicate]

Question

I am slightly confused on the logical NOT operator in Javascript (!). From my understanding, this is mainly used to "inverse" a boolean value. For example, if an expected output of a boolean is true, putting this in front would turn it to false, and vice versa.

In my below code, I created a function allowing user to input a lower and upper integer, and the function would generate a random number between this range. If, however, the user inputs a string instead of an integer, it will prompt the user to enter an integer instead.

I am using isNaN to check if user's input is an integer, and using logical NOT operator in front of it to inverse the result.

In my if condition, if I check isNaN for lower && isNaN for upper are both not a number, this program seems to work correctly. However, if I use ||, it doesn't work as expected, as shown in my code below.

Why is this so? By using OR operator, I am saying if either upper or lower is NaN, then prompt the user to enter a valid integer. Why is it a && and not a || when only one condition needs to be true?

const getNumber = function(lower, upper) { if ( !isNaN(lower) || !isNaN(upper) ) { const number = Math.floor(Math.random() * (upper - lower + 1)) + lower; return number; } else { alert("Please enter a valid integer."); } }; // Call the function and pass it different values console.log( getNumber('six',5) );

NaN does not check if a string is a number. You have to parse it first. Then, the parsing result may be NaN if it could not be parsed — Random
– Random, Commented Mar 13, 2021 at 10:29
@Random - isNaN will use implicit string->number parsing if you pass it a string. (In contrast, Number.isNaN won't.) — T.J. Crowder
– T.J. Crowder, Commented Mar 13, 2021 at 10:29
If you want to use OR you need to remove the NOT - Using isNan here is somewhat of a double negative situation, so you either want x AND y are both NOT isNan, or x OR y isNan — pilchard
– pilchard, Commented Mar 13, 2021 at 10:30

T.J. Crowder · Accepted Answer · 2021-03-13 10:31:21Z

It's not the ! operator that's the problem, it's ||. if ( !isNaN(lower) || !isNaN(upper) ) { says (roughly) "if lower is a number or upper is a number". But you don't wan to say "or" there, because you want them both to be numbers.

So either use use && (and):

if ( !isNaN(lower) && !isNaN(upper) ) { // −−−−−−−−−−−−−−−−^^ const number = Math.floor(Math.random() * (upper - lower + 1)) + lower; return number; } else { alert("Please enter a valid integer."); }

or reverse the content of your if and else blocks and use || without !:

if ( isNaN(lower) || isNaN(upper) ) { alert("Please enter a valid integer."); } else { const number = Math.floor(Math.random() * (upper - lower + 1)) + lower; return number; }

Side note: You're using implicit string-to-number parsing in your code. I recommend doing it on purpose. My answer here goes into your various options for parsing numbers and their pros and cons.

Mr. Hedgehog · Accepted Answer · 2021-03-13 10:34:43Z

By using OR, you are checking that at least one value should not be NaN.

let a = !isNaN(lower); let b = !isNaN(upper);

a and b can be either true or false. When you use ||, you are telling that at least one of this values should be true. If you check Truth table, you will see that OR will be true for this combitations of a and b:

a == true, b == true
a == false, b == true
a == true, b == false

What you want is to check that a == true and b == true simultaneously - so you have to use AND (&&) which will evaluate to true if and only if a == true and b == true.

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