Why Print() Is Printing My String as an Optional

Why is the word Optional being added when I print a string using a dictionary entry as the variable?

Retrieving a value for a given key from a dictionary is always an optional because the key might not exist then the value is nil. Using String Interpolation "\(...)" the Optional is included as literal string.

To avoid the literal Optional(...) in String Interpolation you have to unwrap the optionals preferred in a safe way

if let first = person["first"] as? String, age = person["age"] as? Int {
   print("Your first name is \(first) and you are \(age) years old.")
}

Why is a non-optional value printed as optional?

Inside the function your returning an Int. However the actual signature of your method is Int? meaning it is in fact an optional and you got it wrong!

Basically your method signature is correct. But when you call the function you're getting an optional as the response and must unwrap it.

print([5,15,512,522].challenge37(count: "5")!) // 1

Additionally had you paid close attention you would have noticed that Xcode must gave you a warning (and solutions to solve it)

Expression implicitly coerced from Int? to Any

Xcode gave you the warning because it found out that you're attempting to print an optional and knows that's usually unwanted. Obviously its solution is to unwrap it either through force unwrap or defaulting.

Optional in text view showing when printing

try with if-let statement:

if let result = trans.value(forKey: "page22") {

    page22TextView?.text = result
}

Or try with guard statement:

guard let result = trans.value(forKey: "page22") else { return }
page22TextView?.text = String(describing: result)

Or you can force upwrap it like:

let result = trans.value(forKey: "page22")

if result != nil {

    page22TextView?.text = result! as! String
}

Or you can follow the way suggested by @MrugeshTank below in answers

Why does SWIFT print Optional(...)

Swift has optional types for operations that may fail. An array index like airports["XYZ"] is an example of this. It will fail if the index is not found. This is in lieu of a nil type or exception.

The simplest way to unwrap an optional type is to use an exclamation point, like so: airports["XYZ"]!. This will cause a panic if the value is nil.

Here's some further reading.

You can chain methods on option types in Swift, which will early exit to a nil without calling a method if the lefthand value is nil. It works when you insert a question mark between the value and method like this: airports["XYZ"]?.Method(). Because the value is nil, Method() is never called. This allows you to delay the decision about whether to deal with an optional type, and can clean up your code a bit.

To safely use an optional type without panicking, just provide an alternate path using an if statement.

if let x:String? = airports["XYZ"] {
    println(x!)
} else {
    println("airport not found")
}

Swift Array is printing word optional from my array dont want to force unwrap

If you print out an optional, you will always have Optional(...) in there. The only way to lose that is to unwrap. That's the only solution.

If you are worried about nil values, check for nil and then unwrap.

Printing value of a optional variable includes the word Optional in Swift

Optional chaining is used here:

let age = self.clientDetail?.getAge()

So return of getAge() is optional value. Try optional binding:

if let age = age {
    println("age.....\(age)")
}

or simply unwrap the age with age!, but this will crash your app if age is nil.

Unable to print without Optional()

So, rather than checking whether something is nil, and then when you find out it's not, you both force unwrap and force cast it, we can instead pull this value out safely using optional binding and the correct NSUserDefaults method: stringForKey(_:):

if let data = NSUserDefaults.standardUserDefaults().stringForKey("newString") {
    print(data)
}

But, your second & third tries shouldn't be showing the string as optional. This playground demonstrates effectively the exact same thing as what you're doing here with NSUserDefaults.

func optionalFunc() -> Any? {
    return "Hello world."
}

print(optionalFunc())
/* unsafe */ print(optionalFunc() as! String!) /* unsafe */ 
/* unsafe */ print(optionalFunc() as! String)  /* unsafe */ 
/* unsafe */ print(optionalFunc()!)            /* unsafe */ 

if let unwrapped = optionalFunc() as? String {
    print(unwrapped)
}

As you can see, only the first case prints the Optional("Hello world.") string.

---

Related Topics