In this playground example I'm hoping to find an extension that will remove any nils I put into the creation of a dictionary.
var someValue: String?
if false {
someValue = "test"
}
var dict = ["key": "value",
"key2": someValue]
print("\(dict)")
dict["key3"] = nil
print("\(dict)")
In the above code the current log is
[AnyHashable("key2"): nil, AnyHashable("key"): Optional("value")]
[AnyHashable("key2"): nil, AnyHashable("key"): Optional("value")]
key3 is never added because setting = nil tells it to be removed. I would like to add that functionality to the initial creation of the dictionary but have yet to find a solution that works.
A working solution would result in the following print out
[AnyHashable("key"): Optional("value")]
[AnyHashable("key"): Optional("value")]
This is not solvable in Swift. The correct way to write it is:
var dict = ["key": "value"]
if false {
dict["key2"] = "test"
}
Swift doesn't provide the kind of syntax you're describing, and trying to force it to is going to break the type and create buggy situations. Do not try to create [AnyHashable: Any?]. That is a completely broken type that's going to burn you (Any? is completely broken as a type because Optional is Any, and anything can implicitly become Optional, so it becomes a bizarre recursive rabbit hole). [AnyHashable: Any] is acceptable if you must bridge to NSDictionary, but in general it should be strongly avoided and limited to just where you need it.
Note that this was much even more broken in ObjC (you could write this kind of stuff, but then it'd crash or truncate your dictionary, or some other weird bug), so at least we're making some progress.
In a lot of cases when I see people run into this problem, it's because they've overused optionals in the first place. In your user.name example, why is name optional in the first place? Is there any difference between nil and ""? If there isn't (and there usually isn't), then just make name non-optional (nonnullable in ObjC) and default it to empty and lots of problems go away. Having two versions of the same value (i.e. nil and "" have the same meaning) indicates a type problem, not a syntax problem.
If you want to simplify the syntax just a little bit with a quick extension that works on key/value? pairs like this:
extension Dictionary {
init(keyOptionalPairs: [(Key, Value?)]) {
var d: [Key: Value] = [:]
for (key, value) in keyOptionalPairs {
d[key] = value
}
self = d
}
}
let keyValues: [(String, String?)] = [
("key", "value"),
("key2", nil)
]
let dict = Dictionary(keyOptionalPairs: keyValues)
But notice that the Dictionary is [String: String], not [String: String?]. That's on purpose.
Compared to your syntax, it just adds a set of parens. But compare to the non-fancy version, which isn't beautiful, but is very straightforward.
let dict: [String: String] = {
var d: [String: String] = [:]
d["key"] = "value"
d["key2"] = nil
return d
}()
In Swift The Programming Language Book, I quote:
You can use subscript syntax to remove a key-value pair from a dictionary by assigning a value of nil for that key
They said that assigning a value to nil while remove the pair, but it is not mentioned while initializing the dictionary, so I think it is not valid.
If you really need to do that, I suggest to do some logic after initializing the dictionary to do that for you, like that:
for (key, value) in dict {
if value == nil {
dict[key] = nil
}
}
Related
Here is my swift 2.X code that does'nt work any more on swift 3 :
var dictThemesNamesStyles=[String:[Int:Int]]()
self.styles=dictThemesNamesStyles
let keysArray:Array=Array(self.styles.keys)
let sortedKeysArray = keysArray.sorted(by:
{
(str1: NSObject, str2: NSObject) -> Bool in
return Int((str1 as! String))<Int((str2 as! String))
})
self.stylesLevel1Keys=sortedKeysArray
self.styleThemesPickerView.reloadAllComponents()
On line :
"return Int((str1 as! String)) < Int((str2 as! String))"
it complains with the error : "Type of expression is ambiguous without more context"
What do I have to change in this code to make it work ?
Thanks a lot.
Let's go through this line by line:
var dictThemesNamesStyles=[String:[Int:Int]]()
self.styles=dictThemesNamesStyles
Okay, we've got a dictionary of strings to dictionaries of integers.
let keysArray:Array=Array(self.styles.keys)
There are a few problems with this line:
The declaration of :Array without a generic parameter.
The type declaration is unnecessary, since the type system already knows this is an Array, since you're calling Array's initializer.
Creating this whole Array is unnecessary, since we're just passing the result to sorted, which already exists on the collection returned by keys, and which will return an Array. Creating the array is therefore a needless performance hit which we should avoid.
I would, in fact, delete this entire line, and just replace keysArray with self.styles.keys in the next line:
let sortedKeysArray = self.styles.keys.sorted(by:
Next:
{
(str1: NSObject, str2: NSObject) -> Bool in
return Int((str1 as! String))<Int((str2 as! String))
Okay, we've got a few problems here.
str1 and str2 are declared as NSObject, when they are in fact Strings.
Consequently, the as! String casts are unnecessary.
Int(String) returns an Optional, so you need to take into account the case where the result may be nil. I'd just provide a default value; probably 0, although you could also use a guard statement to throw an error if you prefer.
In general, there's a lot of verbosity here. This whole closure can actually be succinctly written as a one-liner:
let sortedKeysArray = self.styles.keys.sorted { (Int($0) ?? 0) < (Int($1) ?? 0) }
Anyway, take care of these issues, and your code will compile.
That all seems pretty complex. You can simplify this quite a bit:
self.styles = [String: [Int: Int]]()
self.stylesLevel1Keys= self.styles.keys.sorted { Int($0)! < Int($1)! }
self.styleThemesPickerView.reloadAllComponents()
I have below code in my project.
for (key, value) in photoDic {
if let url = URL.init(string: value as! String){
let photo : PhotoRecord = PhotoRecord.init(name:key as! String, url:url)
self.photoRecords.append(photo)
}
}
My question is how can I make key and value in for loop optional, or check if either of them are nil?
I am not able to check if they are nil, getting warning saying any cannot be nil because it is nonoptional.
I was thinking of using something like
for(key:String?, value:String?){}
But it is not working.
The key in a dictionary can't be an optional. (The key must conform to the Hashable protocol, and optionals don't.) So you CAN'T make the keys in your dictionary optional
If you want the values of your dictionary to be Optionals then you need to declare them as Optionals.
So, for example, change
let photoDic: [String: String] = ["key1": "http://www.someDomain.com/image.jpg"]
to
let photoDic: [String: String?] = ["key1": "http://www.someDomain.com/image.jpg"]
(Note that the type of photoDic is changed to [String: String?].)
As mentioned already all keys in a dictionary are non-optional by definition.
Further in NSDictionary all values are non-optional by definition, too.
Be happy about that because
There is no need to check for nil.
The code will never crash.
A Swift dictionary can theoretically contain optional values but practically you are discouraged from using it. For compatibility reasons to NSDictionary a nil value indicates key is missing.
I have a dictionary initialized
var dictionary = [String: [Double]]()
And I want to append
dictionary["Hello"].append(0.0)
but this gives me error "nil".
I tried to solve this by
extension Dictionary {
func appendish(key: String, value: Double) {
if self[key] == nil {
this give me error "Ambiguous reference to member 'subscript'"
}
}
}
How do I solve this? Been stuck at this for hours.
Subscripting a Dictionary with a key returns an optional of type Value?. In your case, dictionary["Hello"] returns a [Double]?. This optionality models the possibility that the dictionary doesn't contain a value for the given key.
If you're only dealing with static data, it's best to just use a literal expression:
let dictionary = [
"Hello": [0.0]
]
If you're using dynamic data, then there are several ways to do what you're trying to achieve, depending on how you would like to handle the nil case:
Use optional chaining
dictionary["Hello"]?.append(0.0)
This appends to the array stored for the key "Hello", but does nothing if there's no such value for that key.
This has the downside of making bugs harder to catch, because the consequence of the silent nil case might not be observed until long after this part of the code has run.
Use force unwrapping
dictionary["Hello"]!.append(0.0)
This appends to the array stored for the key "Hello", but crashes the program if there's no such value for that key.
Unlike optional chaining, this makes it easy to catch the point of failure at runtime. Of course, it comes with the drawback of crashing your program.
Handle the nil case in your own way
if var array = dictionary["Hello"] {
dictionary["Hello"] = nil // This line is a performance optimisation that removes the need for array to be copied
array.append(0.0)
dictionary["Hello"] = array
}
else {
print("No array for the key \"Hello\"") // Handle this as you wish
}
A dictionary look up returns an Optional value because the key might not exist, in which case it returns nil.
If your intention is to append to the array if it exists or create one if there isn't one yet, then the nil coalescing operator ?? comes in handy:
var dict = [String: [Double]]()
dict["hello"] = (dict["hello"] ?? []) + [1]
print(dict) // ["hello": [1.0]]
dict["hello"] = (dict["hello"] ?? []) + [2]
print(dict) // ["hello": [1.0, 2.0]]
This method does create a new array instead of mutating the existing one.
There are a few ways you can do this. Firstly, this is incorrect code:
dictionary["Hello"].append(0.0)
There might not be an array associated with the key "Hello", in which case nil will be returned by the subscript of the dictionary. So you need to unwrap it, either forced or un-forced:
dictionary["Hello"]?.append(0.0)
// or
dictionary["Hello"]!.append(0.0)
But I think what you really want to do is
if dictionary["Hello"] != nil {
dictionary["Hello"]!.append(0.0)
} else {
dictionary["Hello"] = [0.0]
}
After a long time of fiddling around with extensions and stuff (I am not familiar with this area of swift), I finally wrote the method appendish method that you were intended to write:
extension Dictionary where Value : RangeReplaceableCollection & ExpressibleByArrayLiteral, Value.Iterator.Element == Value.Element {
mutating func appendish(key: Key, value: Value.Element) {
if self[key] != nil {
self[key]!.append(value)
} else {
self[key] = [value]
}
}
}
// test
var dict = [String: [Double]]()
dict.appendish(key: "Hello", value: 0.0)
Semantically speaking, [String:AnyObject?] and [String:AnyObject] are the same thing in terms of the way they act, meaning they will return the same thing if I access a key that wasn't set and setting a key to nil will remove that key from the dictionary. Why aren't they considered the same type?
EDIT: I understand the difference from the compiler point of view from the answers. I decided to put the following code in the playground:
var optional = [String:AnyObject?]()
var regular = [String:AnyObject]()
//Some control keys
optional["controlkey"] = "valueoptional"
regular["controlkey"] = "valueRegular"
//Set the keys
optional["keyOptional"] = "valueoptional"
regular["keyRegular"] = "valueRegular"
//Unset the keys
optional["keyOptional"] = nil
regular["keyRegular"] = nil
for (key,val) in optional {
print("key: \(key)\tval: \(val)")
}
for (key,val) in regular {
print("key: \(key)\tval: \(val)")
}
To my surprise the optional did not print the key that was set to nil.
Output was as follow:
key: controlkey val: Optional(valueoptional) //The keys for the optional dictionary
key: controlkey val: valueRegular //The keys for the `regular` dictionary
Why doesn't the key that I set to nil show up?
From the Apple docs:
The Swift language defines the postfix ? as syntactic sugar for the
named type Optional, which is defined in the Swift standard
library.
The type AnyObject? is an enumeration with two cases, None and Some(Wrapped), which are used to represent values that may or may not be present. But AnyObject is responding to one that will be presented.
Setting nil for a key in dictionary means removing the element itself.
Try to set NSNull()
optional["keyOptional"] = NSNull()
Because a nullable type and a non nullable type aren't the same thing from the compiler point of view. It just happens that the Dictionnary interface eventually "blend them" in similar entity, but this is linked with the Dictionnary implementation, not really with the type system.
EDIT: Your update changes the context of the question, but setting to nil is a way to unset from my understanding.
They are different because you can actually store nil into [String:AnyObject?]. You cannot do it using optional[key] = value (subscript operator) because that one has a special behavior for nil (removes value) but you can do it:
var optional: [String:AnyObject?] = ["test": nil]
optional.updateValue(nil, forKey: "test2")
print(optional) // ["test2": nil, "test": nil]
Of course, getting a value from such a dictionary:
print(optional["test"]) // Optional(nil)
results in a double optional Optional<Optional<AnyObject>> (or AnyObject??) and interaction with such types is cumbersome so you should avoid storing nil into dictionaries.
Because setting a key to nil remove that key from the dictionary! From the documentation
Reading a key that is not present in self yields nil. Writing nil as the value for a given key erases that key from self
You can't store nil as a dictionary value. If you need something denote nothingness, use NSNull:
optional["keyOptional"] = NSNull()
regular["keyRegular"] = NSNull()
They aren't the same type because they aren't the same type ;-) Compilers are such literal creatures.
#Sulthan's answer is correct and more complete - I didn't see it until after I hit post on my own answer. But if you look up a key in a [String: AnyObject?] dictionary and a value is present, you'll get an optional wrapped in an optional - the result is an AnyObject??. You'd have to unwrap it twice to use it:
var optional = [String: AnyObject?]()
// I changed it to NSString because String isn't an AnyObject
v
optional["controlkey"] = NSString(string: "valueoptional")
print(optional["controlkey"]) // prints "Optional(Optional(valueoptional))"
if let val = optional["controlkey"] {
print(val) // prints "Optional(valueoptional)"
if let unwrappedVal = val {
print(unwrappedVal) // prints "valueoptional"
}
}
Up until now, I've been unwrapping Optionals in Swift 2.1 like so:
#IBOutlet var commentTextView: UITextView!
if let comment = user["comment"] as? String {
commentTextView.text = comment
}
I never really thought about it, but I think the reason I was doing this was because I was worried that this statement would throw an error if user["comment"] returned something other than a String:
commentTextView.text = user["comment"] as? String
If user["comment"] isn't a String, will the variable on the left of the assignment operator be assigned and throw an error or will the assignment be skipped?
I guess user is in fact a dictionary [String: Any] and what you really do with if let comment = user["comment"] as? String { ... } is not just unwrapping the optional but a conditional type casting (and then unwrapping an optional result of it):
Use the conditional form of the type cast operator (as?) when you are not sure if the downcast will succeed. This form of the operator will always return an optional value, and the value will be nil if the downcast was not possible. This enables you to check for a successful downcast.
Now, to answer your question, if user["comment"] isn't a String then the result will be that commentTextView.text will be assigned nil value, which is bad because its type is String! (implicitly unwrapped optional) about which we hold a promise that it will never be nil. So, yes, there will be an error, an exception actually, but not at the place you would like it to be but at the moment your application will try to access its value assuming that it's not going to be nil.
What you should really do depends on a particular case.
E.g. if you can make user to be a dictionary like [String: String], then you would be able to truly get to unwrapping the optionals and use something like if let comment = user["comment"] { ... }. Or, if you are totally sure that the value for "comment" key will always be there, then you could just do let comment = user["comment"]!.
But if that's not possible then you have to stick with down-casting and the only other thing you can do is to use forced form of it, that is commentTextView.text = user["comment"] as! String. This one at least will produce an exception right at the spot in case if the value at "comment" happens to be not a String but something else.
nil will be assigned to the variable.
If the type of the variable is a non-optional, you'll get a runtime error.
However if user["comment"] is a String you'll get a compiler error about missing ! or ?.
First we need to know of what type the dictionary "user" is.
I assume it is of an unknown type like [String: AnyObject], otherwise why would you try to unwrap it as an String. Let us write a short test to see what happens:
let dict: [String: AnyObject] = ["SomeKey" : 1]
if let y = dict["SomeKey"] as? String {
print(y)
}
You can see clearly that the value of "SomeKey" is an Integer. Trying to unwrap it as an String triggers no error, the "if" statement is just skipped. If an assignment actually happened is hard to prove (maybe by looking at the assembler code) because the variable "y" simply does not exist after the if statement. I assume it will not be created at all.
If the type of the dictionary is known as [String: String] you can omit the try to unwrap it as a String because it's always clear that the type is String.
let dict2: [String: String] = ["SomeKey" : "SomeValue"]
if let y = dict2["WrongKey"] {
// In this case print(y) will not be called because the subscript operator of the dictionary returns nil
print(y)
}
// In this case print(y) will be called because the key is correct
if let y = dict2["SomeKey"] {
print(y)
}