swift - Change the value that is being set in variable's willSet block

Question

I'm trying to sort the array that is being set before setting it but the argument of willSet is immutable and sort mutates the value. How can I overcome this limit?

var files:[File]! = [File]() {
    willSet(newFiles) {
        newFiles.sort { (a:File, b:File) -> Bool in
            return a.created_at > b.created_at
        }
    }
}

To put this question out of my own project context, I made this gist:

class Person {
    var name:String!
    var age:Int!

    init(name:String, age:Int) {
        self.name = name
        self.age = age
    }
}

let scott = Person(name: "Scott", age: 28)
let will = Person(name: "Will", age: 27)
let john = Person(name: "John", age: 32)
let noah = Person(name: "Noah", age: 15)

var sample = [scott,will,john,noah]



var people:[Person] = [Person]() {
    willSet(newPeople) {
        newPeople.sort({ (a:Person, b:Person) -> Bool in
            return a.age > b.age
        })

    }
}

people = sample

people[0]

I get the error stating that newPeople is not mutable and sort is trying to mutate it.

Answer 1

It's not possible to mutate the value inside willSet. If you implement a willSet observer, it is passed the new property value as a constant parameter.

---

What about modifying it to use didSet?

var people:[Person] = [Person]()
{
    didSet
    {
        people.sort({ (a:Person, b:Person) -> Bool in
            return a.age > b.age
        })
    }
}

willSet is called just before the value is stored.
didSet is called immediately after the new value is stored.

You can read more about property observers here https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/Properties.html

You can also write a custom getter and setter like below. But didSet seems more convenient.

var _people = [Person]()

var people: [Person] {
    get {
        return _people
    }
    set(newPeople) {
        _people = newPeople.sorted({ (a:Person, b:Person) -> Bool in
            return a.age > b.age
        })
    }

}