Swift safely unwrapping optinal strings and ints

问题

When I am about to fire my segue for the 2nd view I also send some values like this:

if let aTime = ads[indexPath.row]["unix_t"].int {
    toView.time = aTime
}

if let aTitle = ads[indexPath.row]["title"].string {
    toView.title = aTitle
}

In the second VC I have declared the varibles like:

var time: Int?
var title: String?

and this is how I unwrap the values:

if time != nil {
   timeLabel.text = String(time!)
}

if title != nil {
   titleLabel.text = title!
}

This all works I never get any error caused by unwrapped varibles or nil values. But is there any easier way to do it?

Right now it feels like I am checking too much

回答1:

I can think of three alternatives.

if/let. Very similar to your current option, but you don't have to implicitly unwrap.

if let time = time {
    timeLabel.text = "\(time)"
}

if let title = title {
    titleLabel.text = title
}

You can even unwrap them on the same line. The downside to this is that if one of them is nil, then neither label will be set.

if let time = time, let title = title {
    timeLabel.text = "\(time)"
    titleLabel.text = title
}

guard/let. If these are in a function like setupViews(), then you can one-line your unwrapping like so:

func setupViews() {
    guard let time = time, let title = title else { return }
    timeLabel.text = "\(time)"
    titleLabel.text = title
}

You can use default values and the ?? operator to unwrap quickly.

timeLabel.text = "\(time ?? 0)"
titleLabel.text = title ?? ""

回答2:

You can use the Nil Coalescing Operator too as given in the Docs:

The nil coalescing operator (a ?? b) unwraps an optional a if it contains a value, or returns a default value b if a is nil. The expression a is always of an optional type. The expression b must match the type that is stored inside a.

The nil coalescing operator is shorthand for the code below:

a != nil ? a! : b

回答3:

if let time = time {
    timeLabel.text = "\(time)"
}

if let title = title {
    titleLabel.text = title
}

It's the same as with your JSON

回答4:

Yes, you are checking too much (twice).

Since only non-optional values are passed you can declare the variables non-optional

var time = 0
var title = ""

and set the labels

timeLabel.text = "\(time)"
titleLabel.text = title

The strong type system of Swift highly recommends first to consider

Can I accomplish this without optionals?

and – if there is no other option – then use optionals.

回答5:

You don't need to check nil condition in the first view. Please follow below code:

toView.time = ads[indexPath.row]["unix_t"] as? Int

toView.title = ads[indexPath.row]["title"] as? String

time and title variable in second view are optionals. So, when you assign a value to toView.time and toView.title, it will either assign the value of type time and title respectively or nil.

来源：https://stackoverflow.com/questions/38061226/swift-safely-unwrapping-optinal-strings-and-ints

标签

swift

swifty-json

forced-unwrapping