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JavaScript The Solution

Chris Conwell
Chris Conwell
2,268 Points
Posted by Chris Conwell
Chris Conwell
2,268 Points

A different way?

I was excited to see that My solution is almost exactly like the teacher's solution. However a minor difference. IsNaN(num1,num2) Is this proper?

// declare program variables
var num1;
var num2;
var message;

alert("Let's do some math!"); // announce the program

num1 = prompt("Please type a number"); // collect numeric input
num1 = parseFloat(num1); //convert string to float
num2 = prompt("Please type another number");
num2 = parseFloat(num2);


if ( num2 === 0) { // check for zero
  alert("The second number is 0. You can't divide by zero. Reload and try again.");
}

else if (isNaN(num1,num2)){ // Is not a number?
  alert("At least one of the values you typed is not a number. Reload and try again."); 
} else { 
// build an HTML message
  message = "<h1>Math with the numbers " + num1 + " and " + num2 + "</h1>";
  message += num1 + " + " + num2 + " = " + (num1 + num2);  
  message += "<be>";
  message += num1 + " * " + num2 + " = " + (num1 * num2);
  message += "<be>";
  message += num1 + " / " + num2 + " = " + (num1 / num2);
  message += "<be>";
  message += num1 + " - " + num2 + " = " + (num1 - num2);

// write message to web page
document.write(message);
}

6 Answers

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

The "isNaN" function only tests the first argument, the second one is ignored.

To test two arguments use the "or" operator to combine individual calls :point_right: isNaN(num1) || isNaN(num2)

Tural Rzazade
Tural Rzazade
3,750 Points
Tural Rzazade
Tural Rzazade
3,750 Points

It makes me wonder why isNaN doesn't take OR operator for two or more variables within the same parentheses🤔 isNaN(num1 || num2)

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

Logical operators cannot combine terms like that, only complete expressions as in the example I gave before.

You could, however, write a function that would take any number of arguments and check if any of them were NaN:

function areAnyNaN(...values) {
  for (let val of values)
    if (isNaN(val)) return true;
  return false;
}

So then you could write: areAnyNaN(num1, num2)

Sascha Laufenberg
Sascha Laufenberg
2,961 Points
Sascha Laufenberg
Sascha Laufenberg
2,961 Points

If you do this as a recap and are familiar with functions and loops you can also combine this:

function numOne(){ num1 = prompt("Please type a number"); num1 = parseFloat(num1) if (isNaN(num1)){ numOne() } return num1 }

function numTwo (){ do { num2 = prompt("Please type another number") num2 = parseFloat(num2)} while (num2===0 || isNaN(num2)) return num2 }

numOne() numTwo()

Autumn Rhodes
Autumn Rhodes
925 Points
Autumn Rhodes
Autumn Rhodes
925 Points

I ended up using "else if ( isNaN(num1 + num2) )", which does not seem to have been mentioned, but it works either way.

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

Since doing math on NaN is only one of several reasons an operation might produce NaN, with this test you cannot be certain that either term was NaN to begin with.

Autumn Rhodes
Autumn Rhodes
925 Points
Autumn Rhodes
Autumn Rhodes
925 Points

I'm confused, then, Steven Parker ; when I ran the test using NaN one way or the other, it operated fine. Am I missing something?

Steven Parker
Steven Parker
231,269 Points
Steven Parker
Steven Parker
231,269 Points

Here's an example:

a = "42";
b = Infinity;
console.log("a is " + (isNaN(a) ? "" : "not ") + "NaN...");
console.log("and b is " + (isNaN(b) ? "" : "not ") + "NaN...");
console.log("but a+b is " + (isNaN(a+b) ? "" : "not ") + "NaN!");

a is not NaN..
and b is not NaN..
but a+b is NaN!   :point_left: :astonished:

This difference is writing programs that work most of the time vs. programs that work all of the time.   :wink:

Autumn Rhodes
Autumn Rhodes
925 Points
Autumn Rhodes
Autumn Rhodes
925 Points

Steven Parker AHAH! Blind eyes regardless of surgery. Thank you!