I want to unwrap two optionals in one if statement, but the compiler complaints about an expected expression after operator at the password constant.
What could be the reason?
if let email = self.emailField?.text && let password = self.passwordField?.text
{
//do smthg
}
Done in Swift.
Great news. Unwrapping multiple optionals in a single line is now supported in Swift 1.2 (XCode 6.3 beta, released 2/9/15).
No more tuple/switch pattern matching needed. It's actually very close to your original suggested syntax (thanks for listening, Apple!)
if let email = emailField?.text, password = passwordField?.text {
}
Another nice thing is you can also add where for a "guarding condition":
var email: String? = "baz#bar.com"
var name: String? = "foo"
if let n = name, e = email where contains(e, "#") {
println("name and email exist, email has #")
}
Reference: XCode 6.3 Beta Release Notes
Update for Swift 3:
if let email = emailField?.text, let password = passwordField?.text {
}
each variable must now be preceded by a let keyword
How about wrapping the optionals in a tuple and using switch to pattern match?
switch (self.emailField?.text, self.passwordField?.text) {
case let (.Some(email), .Some(password)):
// unwrapped 'email' and 'password' strings available here
default:
break
}
It's definitely a bit noisier, but at least it could also be combined with a where clause as well.
The usage
if let x = y {
}
is not equivalent to
if (let x = y) { // this is actually not allowed
}
"if let" is effectively a two-word keyword, which is equivalent to
if y != nil {
let x = y!
// rest of if let block
}
Before Swift 1.2
Like #James, I've also created an unwrap function, but this one uses the existing if let for control flow, instead of using a closure:
func unwrap<T1, T2>(optional1: T1?, optional2: T2?) -> (T1, T2)? {
switch (optional1, optional2) {
case let (.Some(value1), .Some(value2)):
return (value1, value2)
default:
return nil
}
}
This can be used like so:
if let (email, password) = unwrap(self.emailField?.text, self.passwordField?.text)
{
// do something
}
From: https://gist.github.com/tomlokhorst/f9a826bf24d16cb5f6a3
Note that if you want to handle more cases (like when one of the two fields is nil), you're better off with a switch statement.
Swift 4
if let suggestions = suggestions, let suggestions1 = suggestions1 {
XCTAssert((suggestions.count > suggestions1.count), "TEST CASE FAILED: suggestion is nil. delete sucessful");
}
I can't explain why the above code doesn't work, but this would be good a replacement:
if let email = self.emailField?.text
{
if let password = self.passwordField?.text
{
//do smthg
}
}
Based on #Joel's answer, I've created a helper method.
func unwrap<T, U>(a:T?, b:U?, handler:((T, U) -> ())?) -> Bool {
switch (a, b) {
case let (.Some(a), .Some(b)):
if handler != nil {
handler!(a, b)
}
return true
default:
return false
}
}
// Usage
unwrap(a, b) {
println("\($0), \($1)")
}
Related
I want to use Swift (not Objective-C runtime) Reflection to create a method like this:
func valueFor(property:String, of object:Any) -> Any? {
...
}
To some extent, I can do this using:
func valueFor(property:String, of object:Any) -> Any? {
let mirror = Mirror(reflecting: object)
return mirror.descendant(property)
}
With
class TestMe {
var x:Int!
}
let t = TestMe()
t.x = 100
let result = valueFor(property: "x", of: t)
print("\(result); \(result!)")
This prints out what I'd expect:
Optional(100); 100
When I do:
let t2 = TestMe()
let result2 = valueFor(property: "x", of: t2)
print("\(result2)")
The output is:
Optional(nil)
This might seem reasonable, except that if I do:
var x:Int!
print("\(x)")
This prints out:
nil
and not Optional(nil). The bottom line is that I'm having difficulty programmatically determining that the value of t2.x is nil using my valueFor method.
If I continue the above code with:
if result2 == Optional(nil)! {
print("Was nil1")
}
if result2 == nil {
print("Was nil2")
}
Neither of these print statements output anything.
When I put a breakpoint into Xcode and look at the value of result2 with the debugger, it shows:
▿ Optional<Any>
- some : nil
So, my question is: How can I determine if the original member variable was nil using the result from valueFor?
Additional1:
If I do:
switch result2 {
case .some(let x):
// HERE
break
default:
break
}
and put a breakpoint at HERE, the value of x turns out to be nil. But, even if I assign it to an Any?, comparing it to nil is not true.
Additional2:
If I do:
switch result2 {
case .some(let x):
let z:Any? = x
print("\(z)")
if z == nil {
print("Was nil3")
}
break
default:
break
}
This prints out (only):
Optional(nil)
I find this especially odd. result2 prints out exactly the same thing!
This is a bit of a hack, but I think it's going to solve the problem for me. I'm still looking for better solutions:
func isNilDescendant(_ any: Any?) -> Bool {
return String(describing: any) == "Optional(nil)"
}
func valueFor(property:String, of object:Any) -> Any? {
let mirror = Mirror(reflecting: object)
if let child = mirror.descendant(property), !isNilDescendant(child) {
return child
}
else {
return nil
}
}
well, i know it has been 4 years, but I am on Xcode 12 and still facing the same issue. since this question seems to be unanswered, I will add what worked for me.
func valueFor(property: String, of object: Any) -> Any? {
let optionalPropertyName = "some"
let mirror = Mirror(reflecting: object)
if let child = mirror.descendant(property) {
if let optionalMirror = Mirror(reflecting: child), optionalMirror.displayStyle == DisplayStyle.optional {
return optionalMirror.descendant(optionalPropertyName)
} else {
return child
}
} else {
return nil
}
}
by using Mirror to check for optional and then extract the optional using "some" you get back either a true object or nil. when this is returned to the caller via the Any? return, you are now able to nil check the value and have that work appropriately.
I am using an SQLite library in which queries return optional values as well as can throw errors. I would like to conditionally unwrap the value, or receive nil if it returns an error. I'm not totally sure how to word this, this code will explain, this is what it looks like:
func getSomething() throws -> Value? {
//example function from library, returns optional or throws errors
}
func myFunctionToGetSpecificDate() -> Date? {
if let specificValue = db!.getSomething() {
let returnedValue = specificValue!
// it says I need to force unwrap specificValue,
// shouldn't it be unwrapped already?
let specificDate = Date.init(timeIntervalSinceReferenceDate: TimeInterval(returnedValue))
return time
} else {
return nil
}
}
Is there a way to avoid having to force unwrap there? Prior to updating to Swift3, I wasn't forced to force unwrap here.
The following is the actual code. Just trying to get the latest timestamp from all entries:
func getLastDateWithData() -> Date? {
if let max = try? db!.scalar(eventTable.select(timestamp.max)){
let time = Date.init(timeIntervalSinceReferenceDate: TimeInterval(max!))
// will max ever be nil here? I don't want to force unwrap!
return time
} else {
return nil
}
}
Update: As of Swift 5, try? applied to an optional expression does not add another level of optionality, so that a “simple” optional binding is sufficient. It succeeds if the function did not throw an error and did not return nil. val is then bound to the unwrapped result:
if let val = try? getSomething() {
// ...
}
(Previous answer for Swift ≤ 4:) If a function throws and returns an optional
func getSomething() throws -> Value? { ... }
then try? getSomething() returns a "double optional" of the
type Value?? and you have to unwrap twice:
if let optval = try? getSomething(), let val = optval {
}
Here the first binding let optval = ... succeeds if the function did
not throw, and the second binding let val = optval succeeds
if the return value is not nil.
This can be shortened with case let pattern matching to
if case let val?? = try? getSomething() {
}
where val?? is a shortcut for .some(.some(val)).
I like Martin's answer but wanted to show another option:
if let value = (try? getSomething()) ?? nil {
}
This has the advantage of working outside of if, guard, or switch statements. The type specifier Any? isn't necessary but just included to show that it returns an optional:
let value: Any? = (try? getSomething()) ?? nil
I have two different scenarios where I need to test the "optionality" of an optional type. I have not been able to figure how to explicitly test if the variable is a .None or a .Some other than with an unwieldy switch statement. How can I test for Someness with an if statement?
Scenario 1
I am writing an address formatter and my inputs are a number of String? types. In this example a simple test for (str != nil) will work. However, since my other need is when dealing with a 'double optional' and a nil test can't distinguish between .Some(.None) and .None a solution to this problem will solve that problem too.
Here's a version that works using a switch
let address1:String? = "123 Main St"
let address2:String? = nil
let apt:String? = "101"
let components = [address1, address2, apt].filter( { (c) -> Bool in
switch c {
case .Some: return true
case .None: return false
}
}).map { return $0! } //Had to map because casting directly to [String] crashes
print(", ".join(components)) //"123 Main St, 101"
What's I'd like to see is something like with an if:
let nice = ["123 Main St", nil, "303"].filter { (c) -> Bool in
return (c == .Some)
}
print(", ".join(nice))
Scenario 2
This is where a nil test won't work. If something is a String?? it can be any of .None, .Some(.None), or .Some(.Some(String)). In my case, the variable is carrying the recordID from an api call which might either be missing entirely (.None), a value (.Some(.Some("ABDEFG")), or explicitly NULL (.Some(.None)).
let teamNoneNone: String?? = .None
let teamSomeNone: String?? = .Some(.None)
let teamSomeSome: String?? = "My favorite local sportsball team"
if teamNoneNone == nil {
print("teamNoneNone is nil but is it .None? We don't know!") //prints
} else {
print("teamNoneNone is not nil")
}
if teamSomeNone == nil {
print("teamSomeNone is nil")
} else {
print("teamSomeNone is not nil but is it .Some(.None)? We don't know!") //prints
}
if teamSomeSome == nil {
print("teamSomeSome is nil but is it .None? We don't know!")
} else {
print("teamSomeSome is not nil but is it .Some(.None) or .Some(.Some())? We don't know!") //prints
}
Via another SO post I found a workaround like this, but it's not very clear what's happening to a casual reader:
if let team: String? = teamSomeNone {
print("teamSomeNone is Some(.None)") //prints
} else {
print("teamSomeNone is .Some(.Some())")
}
if let tests if a value is .None, and if it isn’t, it unwraps it and binds it to a local variable within an if statement.
Using switch with .Some and .None is really a secondary way of handling optionals, if if let doesn’t cut it. But it almost always does, especially now you can do multiple if lets in a single statement, following the latest release of Swift 1.2 to production.
Wanting to filter out the nils in a collection is a common-enough task that Haskell has a standard function for it, called catMaybe. Here’s a version, which I’ll call catSome, that would do the trick in Swift:
func catSome<T>(source: [T?]) -> [T] {
var result: [T] = []
// iterate over the values
for maybe in source {
// if this value isn’t nil, unwrap it
if let value = maybe {
// and append it to the array
result.append(value)
}
}
return result
}
let someStrings: [String?] = ["123 Main St", nil, "101"]
catSome(someStrings) // returns ["123 Main St", "101"]
Doubly-wrapped optionals are a bit of a pain, so the best solution is to avoid them in the first place – often, via use of optional chaining or flatMap.
But if you do find yourself with some, and all you care about is the inner value, you can unwrap them using a double if let:
// later parts of the let can rely on the earlier
if let outer = teamSomeSome, teamName = outer {
println("Fully unwrapped team is \(teamName)")
}
If you want to explicitly know if a double-optional has an inner nil inside an outer value, but isn’t nil itself, you can use if let with a where clause:
if let teamSomeMaybe = teamSomeNone where teamSomeMaybe == nil {
// this will be executed only if it was .Some(.None)
println("SomeNone")
}
The where clause is an extra conditional that can be applied to the unwrapped value.
I currently do this in my code to cope with optionals...
I do a
fetchedResultController.performFetch(nil)
let results = fetchedResultController.fetchedObjects as [doesnotmatter]
// add all items to server that have no uid
for result in results {
let uid = result.valueForKey("uid") as String?
if uid == nil
{
let name = result.valueForKey("name") as String?
let trainingday = result.valueForKey("trainingdayRel") as Trainingdays?
if let trainingday = trainingday
{
let trainingUID = trainingday.valueForKey("uid") as String?
if let trainingUID = trainingUID
{
let urlstring = "http://XXXX/myGym/addTrainingday.php?apikey=XXXXXX&date=\(date)&appid=47334&exerciseUID=\(exerciseUID)"
let sUrl = urlstring.stringByAddingPercentEscapesUsingEncoding(NSASCIIStringEncoding)
let url = NSURL(string: sUrl!)
// save the received uid in our database
if let dictionary = Dictionary<String, AnyObject>.loadJSONFromWeb(url!)
{
trainingday.setValue(uid, forKey: "uid")
}
self.managedObjectContext!.save(nil)
}
}
}
}
Actually I would also need an "else"-clause for each and every "if let" statement. That seems totally terrible code to me! Is there no better way to do this?
Yes, with switch-case and pattern matching you can achieve this:
var x : SomeOptional?
var y : SomeOptional?
switch (x, y)
{
case (.Some(let x), .Some(let y)): doSomething() // x and y present
case (.None(let x), .Some(let y)): doSomethingElse() // x not present, but y
// And so on for the other combinations
default: break
}
Have a look at this blog post: Swift: Unwrapping Multiple Optionals
Edit (slightly off-topic and opinion-based): this is one of my favorite features in Swift. It also lets you implement FSMs with only few code, which is great.
I want to have a constant using let that may be one of several values.
For instance:
if condition1 {
constant = "hi"
}
else if condition2 {
constant = "hello"
}
else if condition3 {
constant = "hey"
}
else if condition4 {
constant = "greetings"
}
I'm not sure how to do this with Swift and the let feature. But I'm inclined to believe it's possible, as this is in the Swift book:
Use let to make a constant and var to make a variable. The value of a constant doesn’t need to be known at compile time, but you must assign it a value exactly once.
How would I accomplish this?
As pointed out in the other answers you can't directly do this. But if you're looking to just variably set the initial value of a constant, then yes, that is possible. Here's an example with a computed property.
class MyClass {
let aConstant: String = {
if something == true {
return "something"
} else {
return "something else"
}
}()
}
I think you are looking for variable which will be assigned later inside switch-case:
let constant :String
switch conditions {
case condition1:
constant = "hi"
case condition2:
constant = "hello"
case condition3:
constant = "hey"
case condition4:
constant = "greetings"
default:
constant = "salute"
}
One option would be something like this, using a closure:
let constant: String = ({ value in
if conditionOne {
return "Hi"
} else if conditionTwo {
return "Bye"
}
return "Oops!"
})(myData /*needed for condition*/)
Or, for another twist, using generics:
func fancySwitch<S, T>(val: S, fn: S -> T) -> T {
return fn(val)
}
let x: String = fancySwitch(3) { val in
if val == 2 {
return "Hi"
} else if val < 5 {
return "Bye"
}
return "Oops"
}
let y: String = fancySwitch((3, 4)) { (a, b) in
if a == 2 {
return "Hi"
} else if b < 5 {
return "Bye"
}
return "Oops"
}
I understand what you're looking for. In Scala and some other functional languages this can be done using the match statement (kind of like switch) because the entire statement resolves to a value like this:
val b = true
val num = b match {
case true => 1
case false => 0
}
This is unfortunately not directly possible in Swift because there is no way to get a value from a branch statement. As stated in the Swift book, "Swift has two branch statements: an if statement and a switch statement." Neither of these statements resolve to a value.
The closest code structure I can think of is to first use a variable to retrieve the correct value and then assign it to a constant to be used in any later code:
let b = true
var num_mutable: Int
switch b {
case true:
num_mutable = 1
default:
num_mutable = 0
}
let num = num_mutable
Just add the line let constant: String before your if/else statement.
Below, an excerpt from Swift 1.2 and Xcode 6.3 beta - Swift Blog - Apple Developer elaborates.
let constants are now more powerful and consistent — The new rule is
that a let constant must be initialized before use (like a var), and
that it may only be initialized, not reassigned or mutated after
initialization. This enables patterns like:
let x : SomeThing
if condition {
x = foo()
} else {
x = bar()
}
use(x)
This formerly required the use of a var even though there is no
mutation taking place. Properties have been folded into this model to
simplify their semantics in initializers as well.
I found the Swift blog post above from the article "Let It Go: Late Initialization of Let in Swift", which I found by googling: swift let constant conditional initialize.