I am using if let for getting the object if its not nil. But I also need to check other condition as well i.e., if "treatmentContext.patientTreatment.canWritePermissions.contains(treatmentContext.pathPatientTreatment.owner". That I am able to do by putting comma after the first statement but here the issue is, I need to unwrap the value of treatmentContext.pathPatientTreatment.owner and here I don't know where exactly I need to unwrap that so that my if condition gets pass when it meets all the criteria.
Below is the code for reference.
if let treatmentContext = IoC.resolve(Treatment.self, from: .treatment), treatmentContext.patientTreatment.canWritePermissions.contains(treatmentContext.pathPatientTreatment.owner)
{
self.presentNavigation(isNew: isNew)
}
You already know you can separate the conditions with ,, so just do that again, but this time it's another optional binding clause:
if let treatmentContext = IoC.resolve(Treatment.self, from: .treatment),
let owner = treatmentContext.pathPatientTreatment.owner,
treatmentContext.patientTreatment.canWritePermissions.contains(owner) {
self.presentNavigation(isNew: isNew)
}
You can separate any number of optional binding clauses, Bool conditions, or case ... patterns with , in an if.
Related
I have an array with 3 elements and want to take the first one and the last one elements.
let array = ["a", "b", "c"]
let first: String = array.first!
let last: String = array.last!
SwiftLint mark a force unwrap as a warning. Can I avoid a forced unwrapping when asking about the first and the last elements for a well known (defined) arrays?
I don't want to use a default values like in an example below
let first :String = array.first ?? ""
Edit:
Why am I asking about it? Because, I would like to avoid an warnings from the SwiftLint when using a forced unwrapping when asking for a first and a last element of an array which was defined by a literal and has enough elements to be sure that there is the first and the last element.
Edit 2:
I have found a name for what I was looking for. It's called Static-Sized Arrays. Static-Sized Arrays discussion stoped in 2017 and there is no chance to use it.
Try with index:
let first = array[0]
let last = array[array.count - 1]
Why am I asking about it? Because, I would like to avoid an warnings
from the SwiftLint when using a forced unwrapping when asking for a
first and a last element of an array which was defined by a literal
and has enough elements to be sure that there is the first and the
last element.
You can't really avoid to unwrap optional value, so if you only need it for two cases extensions can help here.
extension Collection {
func first() -> Element {
guard let first = self.first else {
fatalError() // or maybe return any kind of default value?
}
return first
}
}
let array = [1, 2]
array.first() // 1
And if it need to be only in one swift file you can place this code in that file and mark extensions with private keyword.
Can I avoid a forced unwrapping when asking about the first and the last elements for a well known (defined) arrays?
No you don't have to worry about it for a fixed array , actually the optional attachment for the properties first and last is designated to avoid crashes for an empty arrays
in a practice problem I was asked to print out all elements that are not nil in an array of string, and I realize
for case let name? in names{
print(name)
}
would do the job. But isn't it counter-intuitive?
In the snippet, I read it as "for every element (with actual value or nil)that is in names", but in fact it should be "for every element (actual value)that is in names".
Can anyone help me to make sense of the snippet?
You want to know why this code:
let names = ["b", nil, "x"]
for case let name? in names {
print(name)
}
Produces this output:
b
x
You are wondering what happens to the nil.
The answer is the "Optional Pattern" found in the Language Reference:
The optional pattern provides a convenient way to iterate over an
array of optional values in a for-in statement, executing the body of
the loop only for non-nil elements.
The case keyword is vital. It changes the nature of the for loop significantly. As you can see from this complier error, name? inside the loop is not an optional at all.
Think of the ? as an operator that removes the optionality of name. If the assignment would result in nil, that iteration of the loop does not happen, and the next iteration starts.
Notice that without the case you do not get the same behavior at all.
This:
for name in names {
print(name)
}
Will get this output:
Optional("b")
nil
Optional("x")
And that neither of these work at all.
You can use filter with conditional and then print the result like this:
let nameNotNil = names.filter{$0 != nil} //filter all values diferent nil
print(nameNotNil) // can be nil if any didnt has strings
Suppose I have an array of user's names
let users = ["Hello", "1212", "12", "Bob", "Rob"]
I want to get the first user whose name length is 2, so I filtered the array and got the first user
if let selected = users.filter{$0.characters.count == 2}.first {
print(selected)
}
This code is throwing a compilation error under swift 2.2
Consecutive statements on a line must be separated by ';'
However, this is working fine though
let selected = users.filter{$0.characters.count == 2}.first
if let selected = selected {
print(selected)
}
Can anyone explain why do I need to store filter result in a separate variable first? Any help would be really appreciated.
You can make this work by putting parentheses around the closure that you're passing to filter:
if let selected = users.filter({$0.characters.count == 2}).first {
print(selected)
}
That is the right way to do it. The trailing closure syntax doesn't work very well sometimes on lines with extra elements. You could also put parentheses around the whole statement:
if let selected = (users.filter {$0.characters.count == 2}.first) {
print(selected)
}
Swift is just having trouble parsing your statement. The parentheses give it help in how to parse the line. You should prefer the first way since the closure is indeed a parameter of filter, so enclosing it in parentheses makes it clear to Swift that you are passing it to filter.
I am playing with Arrays in playground and I am bit confused. Here is code:
var players = ["tob", "cindy", "mindy"] //["tob", "cindy", "mindy"]
print(players.isEmpty) // False
var currentPlayer = players.first // "tob"
print(currentPlayer) // "Optional("tob")\n"
Why does it says "Optional"?
I found explanation: "The property first actually returns an optional, because if the array were empty, first would return nil."
But it is not empty. .isEmpty //false, So I am not understanding this.
Thanks for help in advance.
The correct way to think of Optional is that this may or may not have a value. What is the first element of an empty list? There is no such thing. It is not a value. We call that lack of a value nil or .None.
In Swift a variable must have a specific type. So your example:
let currentPlayer = players.first
What is the type of currentPlayer? It may be a String, or it may be nothing at all. It is a "maybe string" and in Swift that's called an Optional<String>. Whether players has elements or doesn't have elements doesn't change the type of currentPlayer.
If you want to do something if-and-only-if the variable has a value, then there are many ways. The simplest is if-let.
let players = ["tob", "cindy", "mindy"] //["tob", "cindy", "mindy"]
print(players.isEmpty) // False
if let currentPlayer = players.first {
print(currentPlayer)
}
This will print tob as you're expecting.
Another very common approach is the guard let
let players = ["tob", "cindy", "mindy"] //["tob", "cindy", "mindy"]
guard let currentPlayer = players.first else { return }
print(currentPlayer)
This lets you avoid nesting the rest of your function inside of curly braces, but otherwise is the same approach.
It is possible to convert an Optional into its underlying type using !, but this is very dangerous and should be avoided except where absolutely necessary. Tools like if-let and guard-let (and also Optional.map) are almost always preferred.
But the key here is to understand that all Swift variables have a single type, and sometimes that type is "maybe it has a value, maybe it doesn't."
If we look at the description of first, we will see that it always returns optional type:
public var first: Self.Generator.Element? { get }
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.