Celsius label isn't applying formula

I have looked through my code like a thousand times, there are no errors detected by Swift and for some reason when I run the application I am getting a bunch of gibberish instead of the celsius degrees. It doesn’t mimic the fahrenheit textfield number like it use to.

Below is my conversionViewController.Swift file

I don’t believe I missed a step in the book and I tried to sort through the code and play around with it, but to no avail. I appreciate your help.

import UIKit

class ConversionViewController: UIViewController {
    
    @IBOutlet var celsiusLabel: UILabel!
    @IBOutlet var textField: UITextField!
    
    var fahrenheitValue: Double? {
        didSet {
            updateCelsiusLabel() //property observer implemented
        }
    }

    
    @IBAction func fahrenheitFieldEditingChanged(textField: UITextField) {
        
        if let text = textField.text, value = Double(text) {
            fahrenheitValue = value
        } else {
            fahrenheitValue = nil
        }
        
    }
    
    @IBAction func dismissKeyboard(sender: AnyObject) {
        textField.resignFirstResponder()
    }
   
    var celsiusValue: Double? {
        if let value = fahrenheitValue {
        return (value - 32) * (5/9)
    } else {
        return nil
    }
}
    func updateCelsiusLabel() {
        if let value = celsiusLabel {
            celsiusLabel.text = "\(value)"
        } else {
            celsiusLabel.text = "???"
        }
        
    }
}

After much strife I finally got it to work. After backtracking and slowly testing the application after every code we have been adding, I noticed that instead of writing “celsiusValue” in the updateCelsiusLabel function, I wrote “celsiusLabel”, which although, it agrees with syntax, it fails to implement the desired change!!

man, talk about trying to debug code, when Swift says there’s no bugs!

It gets worse. You will start getting confusing errors written in Objective-C with no line numbers indicating where the error occurred. Learn to use the debugger as quickly as you can. In your case, you would have observed that celsiusLabel did not contain the value you expected. At the same time, that is an easy error to spot by just reading your code by someone with a little more experience–because of the good, descriptive variable names.