I have noticed when I code in Swift I routinely get an auto correct from the compiler when I use optionals and I am trying to understand the meaning behind this. When I have a variable that is optional and I try to use without unwrapping it, I often get a (?)! autocorrection from Xcode.
In my code I have an optional property that will be the data source collection for my UITableView:
var procedures: [Procedure]?
First I will try to use it like so:
The compiler is telling me that I need to use the ? syntax suffix for my self.procedures.
So I click on the little red circle and have it autocorrect for me like so:
But now the compiler is still complaining. About what you ask? Well, it apparently wants self.procedures?.[indexPath.row] to be enclosed in parenthesis with the ! bang operator at the end...
So I click the little red circle again and have it auto correct like so:
Now the compiler is happy, but I am not. Why am I not happy you ask? Because I do not understand what the () parenthesis are doing here.
Could someone please explain?
So as you can see, you are declaring your procedures as an optional
var procedures: [Procedure]?
then you are trying to pass it to a function that will accept Procedure
cell.configureWithProcedure(procedure: Procedure)
Then you are trying to access your procedure with index like this
self.procedures[indexpath.row]
That will result an error because self.procedures is an optional, so you need to unwrap it, then you try to add ?
self.procedures?[indexpath.row]
This will solve the problem, but the result of it won't be a normal Procedure but an Optional Procedure -> Procedure?.
So XCode give you a hint to unwrap it because your configureWithProcedure ask for Procedure not Procedure? in one line that will be writen like this
(self.procedures?[indexPath.row])!
This operation will result unwrapped Procedure, but I don't recommend you to use the ! operator because it's dangerous, so I recommend you to do like this
if let procedure = self.procedures?[indexPath.row] {
cell.configureWithProcedure(procedure: procedure)
}
EDIT :
in your case you can return an empty cell if procedure isn't exist
extension UITableViewCell {
dynamic class func emptyCell() -> UITableViewCell {
let cell = UITableViewCell(style: UITableViewCellStyle.Default, reuseIdentifier: nil)
cell.selectionStyle = UITableViewCellSelectionStyle.None
return cell
}
}
and write it like this
if let procedure = self.procedures?[indexPath.row] {
cell.configureWithProcedure(procedure: procedure)
} else {
return UITableView.emptyCell()
}
or even better use guard
guard let procedure = self.procedures?[indexPath.row] else {
return UITableView.emptyCell()
}
cell.configureWithProcedure(procedure: procedure)
Hope my answer can help you!
Related
I am asking this hesitantly as I know this is probably a dumb question.
I am returning a Realm result and then have gone ahead and tried to cast it to a String as normal (to put in a text label).
However I'm getting an error 'init' has been renamed to 'init(describing:)'.
When I try use the describing method instead, the label prints "Optional" inside it which obviously isn't what I want.
Is there a reason I can't use :
previousTimeLabel.text = String(lastRecord?.time)
I'm sure I've done this before and it's been fine, am I missing something? (lastRecord.time is an Int).
I've checked the answer here about Interpolation Swift String Interpolation displaying optional? and tried changing to something like this :
if let previousRounds = String(lastRecord?.rounds) {
previousRoundsLabel.text = previousRounds
}
but get the same error + Initializer for conditional binding must have Optional type, not 'String'
The issue isn't String(lastRecord?.time) being Optional. The issue is lastRecord being Optional, so you have to unwrap lastRecord, not the return value of String(lastRecord?.time).
if let lastRecord = lastRecord {
previousRoundsLabel.text = "\(lastRecord.time)"
}
To summarize Dávid Pásztor's answer, here's a way you can fix it:
previousTimeLabel.text = String(lastRecord?.time ?? 0)
This may not be the best way for your application. The point Dávid was making is that you need to deal with lastRecord possibly being nil before trying to pass its time Int into the String initializer. So the above is one simple way to do that, if you're ok with having "0" string as your previousTimeLabel's text if there was no lastRecord.
I am doing the weak strong dance in swift this way:
dispatch_async(dispatch_get_global_queue(QOS_CLASS_DEFAULT, 0), { [weak self] in
guard let `self` = self else {
return
}
self.doSomething(1)
})
Before this, I was using strongSelf instead of `self`. On a website I've seen that I can use this character ` .
But what does this character do in Swift? Without this I cannot assign to self. Why does this work? Is it a good practice to use it?
A bit of an update (I will not refer here when to use it but rather how).
From Swift 4.2 the use should be like:
guard let self = self else { return }
Using ` is basically based on the compiler bug thus not advised.
For more information there is no better source than this. Explaining all reasoning behind and other considerations.
In short the above is more consistent with other cases seen in the code like:
if let myVariable = myVariable
So does not create confusion/discrepancies.
Swift Programming Language
Presents a note that says the following:
If you need to give a constant or variable the same name as a reserved
Swift keyword, surround the keyword with backticks (`) when using it
as a name. However, avoid using keywords as names unless you have
absolutely no choice.
EDIT:
The way I do this is using any other name for example strongSelf like you previously did. Because in the end, both `self` and strongSelf will be some variable to act upon. So I suggest if we can use some other variable name that is fine.
Heys guys,
I am pretty new into programming and therefore I've followed I course on Udemy to teach me Swift 2.2.
For learning purpose I have been trying to program a BMI-calculator where I have a textfield (which just displays the value) and a slider where I can put my weight in kg. After dragging the slider the value is visible in the textfield. I cannot put a value into the textfield so that it is displayed on the slider!
The same textfield-slider relation is used with the height in cm. Now I created an IBAction the bring my kgSlider.value into my kgField.text and it looks like this:
#IBAction func kgSet(sender: AnyObject) {
kgField.text! = String(Int(kgSlider.value))
}
Thats works fine but I unwrapped (like the teacher in the course) without knowing, if there really is a value. Okay, I know in this case that there will be a value, but I would like to go more real and therefore I tried to use an Optional-Binding to find out, if there is a value instead of directly unwrap it.
Therefore I used the cm.Field and the cm.Slider in my code but it doesn't work for now and I don't know why. The code is the following:
#IBAction func cmSet(sender: AnyObject) {
if let tempCm = String(Int(cmSlider.value)) as String! {
cmField.text = tempCm
}
}
So I created the constant called tempCM which will got the value from the cmSlider, if there is a value. Therefore I casted the cmSlider.value like in the other IBAction into an Int and then into a String. If there is the value it will carry it into the cmField.text. This didn't work, therefore I tried to use the "as String!" statement but know I get always 0 instead of the correct value.
So what am I doing wrong there?
So, this should compile fine and provide you with your desired result.
#IBAction func cmSet(sender: AnyObject) {
if let tempCm = String(Int(cmSlider.value)) {
cmField.text = tempCm
}
}
You could also try this
cmField.text = String(Int(cmSlider.value)) ?? " "
in the second example, you are using the optional operator to say if you can convert this to an Int then that Int to a string set the cmField.text property to its value, otherwise use a blank space as the default value.
I'm very new to swift, but proficient in other languages like Java, JavaScript, C, ... I'm lost with Swift syntax when it comes to create expressions. Look at this basic example where I just try to find out if one string is contained into another by calling String.rangeOfString that returns an Optional Range (Range?)
This works as expected:
let LEXEMA:String="http://"
let longUrl:String="http://badgirls.many/picture.png"
let range=longUrl.rangeOfString(LEXEMA);
if (range? != nil) {
// blah
}
Now I'm trying to combine the expression inside the if, something like:
if (longUrl.rangeOfString(LEXEMA)? !=nil) {
// blah
}
But I always get syntax errors, the above yields a "Expected Separator" and can't understand why. Done some more tests:
if (absolutePath.rangeOfString(URL_LEXEMA) !=nil) { }
Expected Separator before "!"
if absolutePath.rangeOfString(URL_LEXEMA) !=nil { }
Braced block of statements is an unused closure
What am I doing wrong?
If you’re coming from other like Java, you might be thinking of optionals like pointers/references, and so used to equating them to nil and if non-nil, using them. But this is probably going to lead to more confusion. Instead, think of them like a container for a possible result, that you need to unwrap to use. if let combines the test and unwrapping operation.
With this in mind, here’s how you could adapt your code:
let LEXEMA: String="http://"
let longUrl: String="http://badgirls.many/picture.png"
if let range = longUrl.rangeOfString(LEXEMA) {
// use range, which will be the unwrapped non-optional range
}
else {
// no such range, perhaps log an error if this shouldn’t happen
}
Note, that ? suffixing behaviour you were using changes in Swift 1.2 so even the code in your question that compiles in 1.1 won’t in 1.2.
It’s possible that sometimes you are whether there was a value returned, but you don’t actually need that value, just to know it wasn’t nil. In that case, you can compare the value to nil without the let:
if longUrl.rangeOfString(LEXEMA) != nil {
// there was a value, but you don't care what that value was
}
That said, the above is probably better expressed as:
if longUrl.hasPrefix(LEXEMA) { }
For starters:
You don't need parenthesis with if statements unless you have nested parenthetical subexpressions that require it.
You don't need to specify the type on the left side of the = of a let or var declaration if Swift can figure it out from the right side of the =. Very often Swift can figure it out, and you can tell that Swift can figure it out, so you can avoid that redundant clutter.
You do need to specify the type if Swift cannot figure out the type from
the right side. Example:
For example, consider the following lines:
let LEXEMA = "http://"
let longUrl = "http://badgirls.many/picture.png"
Swift can figure out that they're strings.
Similarly for this function or class that returns a UIView:
var myView = ViewReturningClassOrFunc()
Consider this:
#IBOutlet var myView : UIView!
In the above line, Swift cannot figure out ahead of time it will be assigned a UIView, so you have to provide the type. By providing a ! at the end you've made it an implicitly unwrapped optional. That means, like ?, you're indicating that it can be nil, but that you are confident it will never be nil at the time you access it, so Swift won't require you to put a ! after it when you reference it. That trick is a time saver and big convenience.
You should NOT add the ? to the line:
if (longUrl.rangeOfString(URL_LEXEMA) !=nil) {
As another answer pointed out, you're missing the let.
if let longUrl.rangeOfString(URL_LEXEMA) {
println("What do I win? :-)")
}
swift is case sensitive language. you need to check about whitespaces as well
if longUrl.rangeOfString(LEXEMA) != nil {
//your condition
}
there should be space between statement != nil
Just add a space between != and nil like:
if longUrl.rangeOfString(LEXEMA) != nil {
// blah
}
I tested your code in playground, an error of Expected ',' separator reported.
And do not forget the rules that 1s and 0s and Airspeed Velocity said.
I am reading the following book, on Page #32 there is a code snippet. December 2014: First Edition.
Swift Development with CocoaJonathon Manning, Paris Buttfield-Addison,
and Tim Nugent
I know we can use ? to make a vairbale optional and ! to unwrap a optional vairbale in Swift
var str: String? = "string"
if let theStr = str? {
println("\(theStr)")
} else {
println("nil")
}
Why do we need ? in this line if let theStr = str? It seems working just fine with out it and making no difference.
You don't need it, and shouldn't have it. The optional binding evaluates the optional. If it's nil, it stops. If it's not nil, it assigns the value to your required variable and then executes the code inside the braces of the if statement.
EDIT:
The language has changed slightly since it was first given out in beta form. My guess is that the ? used to be required.
Some of the sample projects I've used from Github fail to compile and I've had to edit them to get them to work. this might be an example where the syntax has changed slightly.
The current version of Swift does not require it, as it is redundant since your variable is already an optional.
Whatever you put in the if let statement does have to be an optional though or you will receive an error
Bound Value in a conditional binding must be of Optional type
Furthermore, if you are casting to a type, you do need to use as? to cast to an optional type.
var str2: Any = ["some", "example"]
if let theStr = str2 as? [String] {
println("\(theStr)")
} else {
println("nil")
}