Value of Optional Type URL? Must Be Unwrapped to a Value of Type URL

Optionals are an integral aspect of Swift. Optional types are used to represent the absence or presence of a value. In other words, an optional type can hold either a value or nil, which indicates that a value is absent. When you declare a variable as optional, you tell the Swift compiler that the variable may not always have a value.

To work with the actual value contained in an optional, you must unwrap it to access the underlying value. That brings us to the error you are seeing.

value of optional type URL?' must be unwrapped to a value of type URL

The compiler tells you that you try to use an optional value without unwrapping it first to access the underlying value. Let's look at an example to better understand the issue and how to resolve it. Create a playground in Xcode if you'd like to follow along with me. Add an import statement for the Foundation framework at the top. Define a constant urlAsString that stores a URL as a string.

import Foundation

let urlAsString = "https://cocoacasts.com/"

We pass the value stored in urlAsString to one of the initializers of the URL struct to create a URL object. The initializer is failable so the url constant is of type URL?. That is important to understand. If the value stored in urlAsString isn't a valid URL, then the initialization of the URL object fails.

import Foundation

let urlAsString = "https://cocoacasts.com/"
let url = URL(string: urlAsString)

Let's use the URL object to create a URLRequest object. We pass the value stored in the url constant to one of the initializers of the URLRequest struct. The result is a compile-time error. The compiler notifies us that the url constant is of type URL?, an optional type. The initializer of the URLRequest struct expects a URL object, not an optional URL object.

import Foundation

let urlAsString = "https://cocoacasts.com/"
let url = URL(string: urlAsString)

let request = URLRequest(url: url)

The solution is simple, though. We use optional binding to safely access the value stored in the URL constant. In other words, we safely check if the optional contains a value. If it does, then we can use that value to create a URLRequest object. If it doesn't, we print a message to the console.

import Foundation

let urlAsString = "https://cocoacasts.com/"
let url = URL(string: urlAsString)

if let url {
    let request = URLRequest(url: url)
} else {
    print("Invalid URL")
}

You could force unwrap the optional using the exclamation mark, but that isn't something I recommend. Force unwrapping should be done very, very sparingly. Why is that? If you force unwrap an optional that doesn't contain a value, a fatal error is thrown and the process terminates immediately. This simply means that your app crashes.

There are valid use cases for using the exclamation mark to force unwrap optionals, but decide for yourself in advance what those use cases are. Understand the risk involved and stick to those use cases. That is the approach I recommend.