swift code is not working after clean build.
how can i rewrite this simple code?
if let name: String = String(cString: (interface?.ifa_name)!), name == "utun0" {
print("Ipfound")
ipFound = true;
}
i put image for better understanding.
The if let construct is used for "conditional binding" of an optional. The thing you are trying to assign to name (the function result) is not optional - instead, the value you are passing to String(cString:,name:) is optional.
You should rewrite your code to something like this:
if let interfaceName = interface?.ifa_name {
let name = String(cString: interfaceName)
if name == "utun0" {
print("Ipfound")
ipFound = true
} else {
print("name != Ipfound")
}
} else {
print("interface or interface.ifa_name is nil")
}
In that code I'm using optional binding to try to get a non-nil value from interface?.ifa_name.
If you use ! "force unwrap" operator as you were trying to do, your code would crash if interface is nil, or if interface.ifa_name is nil.
Edit:
Another way you could handle this, that would look closer to your original code, would be to use map() on the Optional:
if let name: String = interface?.ifa_name.map ({ String(cString: $0)}),
name == "utun0" }
{
print("Ipfound")
ipFound = true;
}
The form of map() that works on an Optional takes an Optional as input. If that optional is nil, it returns nil. If the optional is not nil, the result returned by map() is the result of applying your closure to the unwrapped optional.
So i've got this read only property:
private var allProducts: [String]? {
if let selectedProductSeries = selectedProductSeries {
return ["someProduct"]
} else {
return nil
}
}
Great, that does what I want, but my question is why do I have to be exhaustive in my if/else. I'm marking it as optional and thus what I'd expect to happen is that if I don't return anything its just nil. But of course if I get rid of the else it complains saying: Missing return in function expected to return: [String]? Which is funny because the message says its expecting me to return something even though its optional.
Why is this?
I would assume that errors might occur more often, if the return nil was implicitly added. Inside of a function/method which contains many if- or switch-statements, you could easily overlook, that you might fall through each and every statement. An explicit return nil forces you to deal with that possibility.
I am getting error:
initialiser for conditional binding must have optional type, not '()'.
Using Swift language, below is the code:
if let result = brain.performOperation(operation)
I think I can answer your question, but I don't know how much it will help you.
The way if let someVar = someOptional { } is used is to check the value of a variable (someOptional) to see if it is nil or a "value". If it is not nil then someVar will be assigned the non-nil value of someOptional and execute the code between { } which can safely reference someVar, knowing it is not nil. If someOptional is nil then the code within { } is bypassed and not executed.
The comment you posted above indicates that the performOperation() method is this:
func performOperation(symbol:String) {
if let operation = knownOps[symbol] {
opStack.append(operation)
}
}
This method does not return anything, or more formally it returns void aka (). void, or () is not a value, nor is it nil.
So when you have this statement
if let result = brain.performOperation(operation) { }
the compiler complains because it expects brain.performOperation(operation)
to return either nil or a value, but not void aka () which is exactly what the method returns.
If you are still confused about optionals, be sure to read as much as you can of the Swift Language Reference. Optionals are a big part of the language and once you get used to them you will find them very invaluable.
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.
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."
}
To remove it, there are three methods you could employ.
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:
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.
The Implicitly Unwrapped Optional
if let unwrappedAge = age {
// continue in here
}
Note that the unwrapped type is now Int, rather than Int?.
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!
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
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
To unwrap optional use age! instead of age. Currently your are printing optional value that could be nil. Thats why it wrapped with Optional.
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.
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
}
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
Check out the guard statement:
for student in class {
guard let age = student.age else {
continue
}
// do something with age
}
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")
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.
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
If you just want to get rid of strings like Optional(xxx) and instead get xxx or nil when you print some values somewhere (like logs), you can add the following extension to your code:
extension Optional {
var orNil: String {
if self == nil {
return "nil"
}
return "\(self!)"
}
}
Then the following code:
var x: Int?
print("x is \(x.orNil)")
x = 10
print("x is \(x.orNil)")
will give you:
x is nil
x is 10
PS. Property naming (orNil) is obviously not the best, but I can't come up with something more clear.
With the following code you can print it or print some default value. That's what XCode generally recommend I think
var someString: String?
print("Some string is \(someString ?? String("Some default"))")
If you are printing some optional which is not directly printable but has a 'to-printable' type method, such as UUID, you can do something like this:
print("value is: \(myOptionalUUID?.uuidString ?? "nil")")
eg
let uuid1 : UUID? = nil
let uuid2 : UUID? = UUID.init()
print("uuid1: \(uuid1?.uuidString ?? "nil")")
print("uuid2: \(uuid2?.uuidString ?? "nil")")
-->
uuid1: nil
uuid2: 0576137D-C6E6-4804-848E-7B4011B40C11
I'm going through the swift docs, and in the optional segment, it talks about using the question mark -- ? -- to signify variables that might be nil. This can be used in an if statement to check for nil, but in the docs they assign the optional to a new variable in the conditional. Is there a reason for this?
For Example, it is presented in the docs similar to this:
// Declare an optional string (might be nil)
var optionalString: String? = "Hello"
// Assigns optionalString to new variable before checking if nil
if let string = optionalString {
println("\(optionalString) is not nil!")
}
else {
println("\(optionalString) is nil")
}
However, this runs just fine for me in tests:
var optionalString: String? = "Hello"
// Assigns optionalString to new variable before checking if nil
if optionalString {
println("\(optionalString) is not nil!")
}
else {
println("\(optionalString) is nil")
}
Question
Is there a reason to assign optionalString to a new variable string in the conditional statement?
Take a look at the section on Optional Chaining in the docs. In the example you cite, there's not much difference. But in other cases, an if-let construction lets you get at an unwrapped value that comes from a series of optional references and method calls, without using implicit unwraps that can crash your app if you haven't considered all the possible bindings for a value in a chain.
It's also useful if you want to avoid recomputing a value. You can use it in a lot of the same ways you'd use an assignment in a conditional in (Obj)C (remember if (self = [super init])).
For example, if the optional being tested comes from a computed property:
var optionalName: String? {
get {
if checkTouchID() {
return "John Appleseed"
} else {
return nil
}
}
}
var greeting = "Hello!"
if optionalName != nil {
greeting = "Hello, \(optionalName)"
}
Paste that into a playground, along with a stub implementation of checkTouchID() that returns true, and you'll immediately see in the results area that the optionalName getter is executing twice. (This would be a problem in a more realistic scenario, because you probably don't want code like this to implicitly checkTouchID() or downloadFromServer() or billApplePay() twice.) If you use an if-let construction instead, you'll only execute the getter once.
In a series of chained optionals (like if let johnsStreet = john.residence?.address?.street in the docs linked above), you don't want to rewrite the whole chain in the body of the if statement, much less recompute it.
I think the purpose of that assignment was to demonstrate the use of "let" within the if conditional clause. I don't see a meaningful difference between the provided code and your own.
From: Apple Inc. “The Swift Programming Language.” iBooks. https://itun.es/il/jEUH0.l
“If the optional value is nil, the conditional is false and the code in braces is skipped. Otherwise, the optional value is unwrapped and assigned to the constant after let, which makes the unwrapped value available inside the block of code.”