Finding a type of a variable in JS

Question

I have a following task from my mentor:

1. Type a value with a prompt and make it a number with unary plus

2. console.log if it's even or odd

3. If it's not a number, add another console log

Here's the code I have right now, and I can't really understand what's wrong:

x = prompt("Enter your value: ");

if(x % 2 === 0) {
    console.log("x is an even number");
} else  {
    console.log("x is an odd number")
}

if (typeof x === "!Number"){
    console.log("Whoops, it seems like you're mistaken");
} 

Answer 1

There're several issues in your code:

1. You should check the type of x before you judge it's even or odd.

2. The type of x can only be string since it's come from prompt, but you can still know if it's a valid number string by isNaN() function.

3. If you want to know whether a type of a variable is number or not, it should be typeof x !== 'number'. typeof x === '!Number' will never be true(also the case matters).

Here's an example that how you could write the code:

let x = prompt("Enter your value: ");

if (isNaN(x)){
    console.log("Whoops, it seems like you're mistaken");
} else if(x % 2 === 0) {
    console.log("x is an even number");
} else  {
    console.log("x is an odd number")
}