swift - Printing optional variable


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I am trying with these lines of code

class Student {
    var name: String
    var age: Int?

    init(name: String) {
        self.name = name
    }

    func description() -> String {
        return age != nil ? "\(name) is \(age) years old." : "\(name) hides his age."
    }
}

var me = Student(name: "Daniel")
println(me.description())
me.age = 18
println(me.description())

Above code produces as follow

Daniel hides his age.
Daniel is Optional(18) years old.

My question is why there is Optional (18) there, how can I remove the optional and just printing

Daniel is 18 years old.


All Answers
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    You have to understand what an Optional really is. Many Swift beginners think var age: Int? means that age is an Int which may or may not have a value. But it means that age is an Optional which may or may not hold an Int.

    Inside your description() function you don't print the Int, but instead you print the Optional. If you want to print the Int you have to unwrap the Optional. You can use "optional binding" to unwrap an Optional:

    if let a = age {
     // a is an Int
    }
    

    If you are sure that the Optional holds an object, you can use "forced unwrapping":

    let a = age!
    

    Or in your example, since you already have a test for nil in the description function, you can just change it to:

    func description() -> String {
        return age != nil ? "\(name) is \(age!) years old." : "\(name) hides his age."
    }
    

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    An optional means that Swift is not entirely sure if the value corresponds to the type: for example, Int? means that Swift is not entirely sure whether the number is an Int.

    To remove it, there are three methods you could employ.

    1) If you are absolutely sure of the type, you can use an exclamation mark to force unwrap it, like this:

    // Here is an optional variable:
    
    var age: Int?
    
    // Here is how you would force unwrap it:
    
    var unwrappedAge = age!
    

    If you do force unwrap an optional and it is equal to nil, you may encounter this crash error:

    enter image description here

    This is not necessarily safe, so here's a method that might prevent crashing in case you are not certain of the type and value:

    Methods 2 and three safeguard against this problem.

    2) The Implicitly Unwrapped Optional

     if let unwrappedAge = age {
    
     // continue in here
    
     }
    

    Note that the unwrapped type is now Int, rather than Int?.

    3) The guard statement

     guard let unwrappedAge = age else { 
       // continue in here
     }
    

    From here, you can go ahead and use the unwrapped variable. Make sure only to force unwrap (with an !), if you are sure of the type of the variable.

    Good luck with your project!


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    For testing/debugging purposes I often want to output optionals as strings without always having to test for nil values, so I created a custom operator.

    I improved things even further after reading this answer in another question.

    fileprivate protocol _Optional {
        func unwrappedString() -> String
    }
    
    extension Optional: _Optional {
        fileprivate func unwrappedString() -> String {
            switch self {
            case .some(let wrapped as _Optional): return wrapped.unwrappedString()
            case .some(let wrapped): return String(describing: wrapped)
            case .none: return String(describing: self)
            }
        }
    }
    
    postfix operator ~? { }
    public postfix func ~? <X> (x: X?) -> String {
        return x.unwrappedString
    }
    

    Obviously the operator (and its attributes) can be tweaked to your liking, or you could make it a function instead. Anyway, this enables you to write simple code like this:

    var d: Double? = 12.34
    print(d)     // Optional(12.34)
    print(d~?)   // 12.34
    d = nil
    print(d~?)   // nil
    

    Integrating the other guy's protocol idea made it so this even works with nested optionals, which often occur when using optional chaining. For example:

    let i: Int??? = 5
    print(i)              // Optional(Optional(Optional(5)))
    print("i: \(i~?)")    // i: 5
    

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    Update

    Simply use me.age ?? "Unknown age!". It works in 3.0.2.

    Old Answer

    Without force unwrapping (no mach signal/crash if nil) another nice way of doing this would be:

    (result["ip"] ?? "unavailable").description.

    result["ip"] ?? "unavailable" should have work too, but it doesn't, not in 2.2 at least

    Of course, replace "unavailable" with whatever suits you: "nil", "not found" etc


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    To unwrap optional use age! instead of age. Currently your are printing optional value that could be nil. Thats why it wrapped with Optional.


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    In swift Optional is something which can be nil in some cases. If you are 100% sure that a variable will have some value always and will not return nil the add ! with the variable to force unwrap it.

    In other case if you are not much sure of value then add an if let block or guard to make sure that value exists otherwise it can result in a crash.

    For if let block :

    if let abc = any_variable {
     // do anything you want with 'abc' variable no need to force unwrap now.
    }
    

    For guard statement :

    guard is a conditional structure to return control if condition is not met.

    I prefer to use guard over if let block in many situations as it allows us to return the function if a particular value does not exist. Like when there is a function where a variable is integral to exist, we can check for it in guard statement and return of it does not exist. i-e;

    guard let abc = any_variable else { return }
    

    We if variable exists the we can use 'abc' in the function outside guard scope.


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    age is optional type: Optional<Int> so if you compare it to nil it returns false every time if it has a value or if it hasn't. You need to unwrap the optional to get the value.

    In your example you don't know is it contains any value so you can use this instead:

    if let myAge = age {
        // there is a value and it's currently undraped and is stored in a constant
    }
    else {
       // no value
    }
    

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    I did this to print the value of string (property) from another view controller.

    ViewController.swift

    var testString:NSString = "I am iOS Developer"
    

    SecondViewController.swift

    var obj:ViewController? = ViewController(nibName: "ViewController", bundle: nil)
    print("The Value of String is \(obj!.testString)")
    

    Result :

    The Value of String is I am iOS Developer
    

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    When having a default value:

    print("\(name) is \(age ?? 0) years old")

    or when the name is optional:

    print("\(name ?? "unknown") is \(age) years old")


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    Check out the guard statement:

    for student in class {
        guard let age = student.age else { 
            continue 
         }
        // do something with age
    }
    

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    I was getting the Optional("String") in my tableview cells.

    The first answer is great. And helped me figure it out. Here is what I did, to help the rookies out there like me.

    Since I am creating an array in my custom object, I know that it will always have items in the first position, so I can force unwrap it into another variable. Then use that variable to print, or in my case, set to the tableview cell text.

    let description = workout.listOfStrings.first!
    cell.textLabel?.text = description
    

    Seems so simple now, but took me a while to figure out.


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    This is not the exact answer to this question, but one reason for this kind of issue. In my case, I was not able to remove Optional from a String with "if let" and "guard let".

    So use AnyObject instead of Any to remove optional from a string in swift.

    Please refer link for the answer.

    https://stackoverflow.com/a/51356716/8334818