In my model, I have some arrays:
var thisArray = [Object]
var thatArray = [Object]
var anotherArray = [Object]
In my view controller, I want to switch on a value to determine which array I will append to:
var whichArray: [Object]!
switch someValue {
case .thisArray: whichArray = thisArray
case .thatArray: whichArray = thatArray // "He went thatArray!"
case .anotherArray: whichArray = anotherArray
}
whichArray.append(object)
But of course this won't work because Array is a value type.
Is there a way to do this? Of course I could do the following:
switch someValue {
case .thisArray: thisArray.append(object)
case .thatArray: thatArray.append(object)
case .anotherArray: anotherArray.append(object)
}
But that is so inelegant and redundant! And if there's other more complex things going on in the surrounding code, then it's especially so.
Is there a solution here? Is it possible to create a reference to a value type?
PS. Even better, though really its own question, is if I could use the name of the case (e.g., "thisArray" for someValue = .thisArray) to set the array, by name (i.e., avoid the whole switch statement and just say objectName.append(object) or something like that) but as far as I know this isn't a thing. Or maybe this IS possible? And maybe it's my birthday?
Since Arrays are value types - as you have said yourself - they can't be passed around (or assigned) as a reference. One solution would be to create a wrapper class for the Array which itself would then be a reference type. You can then assign this wrapper class instead of the arrays themselves.
Now, given that you also said you might prefer to access the Arrays by the name and completely get rid of the switch you could change your design to storing thisArray, thatArray and anotherArray in a Dictionary, with the keys being the different values for someValue.
This way you could simply append to the desired array with:
arrayDict[someValue]?.append(object)
(Given that you've properly set up the dictionary beforehand)
Like this for example:
enum Value {
case thisArray
case thatArray
case anotherArray
}
var arrayDict = [
Value.thisArray : [String](),
Value.thatArray : [String](),
Value.anotherArray : [String]()
]
arrayDict[.thatArray]?.append("Some String.")
For the sake of creating a short working example I've replaced Object with String but that obviously doesn't matter.
I would typically recommend solving this with closures. It's more powerful and safer. For example:
let append: (Object) -> Void
switch someValue {
case .thisArray: append = { thisArray.append($0) }
case .thatArray: append = { thatArray.append($0) }
case .anotherArray: append = { anotherArray.append($0) }
}
append(object)
(It would be ideal here to just say append = thisArray.append, but you can't do that in Swift today. It's a "partial application of a mutating function" and that's not currently legal.)
Even though Swift was designed to reduce pointer operations, pointers are still available:
var thisArray = [1,2,3]
var thatArray = [4,5,6]
var anotherArray = [7,8,9]
var ptr: UnsafeMutablePointer<[Int]>
let someValue = 2
switch someValue {
case 1: ptr = UnsafeMutablePointer(&thisArray)
case 2: ptr = UnsafeMutablePointer(&thatArray)
default: ptr = UnsafeMutablePointer(&anotherArray)
}
ptr.pointee.append(42)
print(thatArray) // [4,5,6,42]
A minor annoyance with this is that you have to call ptr.pointee to access the target array. If you assign the pointee to another variable (i.e. let whichArray = ptr.pointee), any modification to whichArray won't be reflected in the original array.
(I had to change your Object type to Int so that it runs in the IBM Swift Sandbox)
Related
I want to initialize every time a struct with dictionaries. Later, I'm going to use its properties instead a dictionary's keys and values - it seems rather easier. However, when I try the code below, it tells me that "Return from initializer without initializing all stored properties" and "1. 'self.one' not initialized" and "2. 'self.two' not initialized". My question is how to initialize a struct from a dictionary, so that I have basically a struct with the contents of the dictionary? Or how to transform it into struct?
struct Blabla {
var one: String
var two: [Int]
init(three: [String: [Int]]) {
for i in three {
self.one = i.key
self.two = i.value
}
} ERROR! - Return from initializer without initializing all stored properties
}
struct Blabla {
var one: String
var two: [Int]
init(three: [String: [Int]]) {
one = ""
two = []
for i in three {
self.one = i.key
self.two = i.value
}
} ERROR! - Return from initializer without initializing all stored properties
}
for in clause may have zero runs, in which case struct properties will not be initialized. You have to provide default values (or emit fatalError if you really need to).
While I think your example is pure synthetical, there is no need to loop through array, you can set properties to its last entry.
The issues is that if three is an empty Dictionary, the instance properties one and two don't get initialised. Also, you are overwriting the properties in each iteration of the for loop and the compiler cannot guarantee that there will be any iterations of the loop in compile-time, hence the compiler error.
You could make the initialiser failable to account for this by checking that the dictionary actually contains at least one key-value pair and assigning that first key-value pair to your properties.
struct Blabla {
var one: String
var two: [Int]
init?(three: [String: [Int]]) {
guard let key = three.keys.first, let value = three[key] else { return nil }
one = key
two = value
}
}
However, you should rethink what it is that you are actually trying to achieve, since with your current setup you have a mismatch between your init input values and the properties of your struct.
This code should compile, but it feels unsafe to me to initialize a Struct in this way because:
It assume your dictionary has values in it.
Your stored properties will always have the last value you looped through.
In order to pull values out to satisfy the compiler you need to force unwrap them. (With Dávid Pásztor's guard-letting approach, this can be avoided)
struct Blabla {
var one: String
var two: [Int]
init(three: [String: [Int]]) {
self.one = three.keys.first!
self.two = three[three.keys.first!]!
}
}
let input = ["pizza": [1,2]]
let l = Blabla(three: input)
If I were you I would let the memberwise initializer that you get for free do its thing and provide either a specialized initializer to handle your case of taking a Dictionary as input or move that parsing to another function/class/etc....
The compiler error is clear: If the dictionary is empty the struct members are never initialized. But the code makes no sense anyway as each iteration of the dictionary overwrites the values.
Maybe you mean to map the dictionary to an array of the struct
struct Blabla {
let one: String
let two: [Int]
}
let three = ["A":[1,2], "B":[3,4]]
let blabla = three.map{Blabla(one: $0.key, two: $0.value)}
print(blabla) // [Blabla(one: "A", two: [1, 2]), Blabla(one: "B", two: [3, 4])]
struct blabla{
var a : string
var b : [int] = []
init(_ data: [string:[int]]){
// whatever you want to do
}
}
Consider the following code:
struct Card {
var name0: String
var name1: String
}
var cards = [Card]()
cards.append(Card(name0: "hello", name1: "world"))
// Need to perform array index access,
// every time I want to mutate a struct property :(
cards[0].name0 = "good"
cards[0].name1 = "bye"
// ...
// ...
// "good bye"
print(cards[0].name0 + " " + cards[0].name1)
Instead of having to perform multiple array index accessing every time I want to mutate a property in struct, is there a technique to avoid such repeating array index accessing operation?
// Ok. This is an invalid Swift statement.
var referenceToCardStruct = &(cards[0])
referenceToCardStruct.name0 = "good"
referenceToCardStruct.name1 = "bye"
// ...
// ...
There are a lot of good answers here, and you should not think of value types as "a limitation." The behavior of value types is very intentional and is an important feature. Generally, I'd recommend inout for this problem, like matt suggests.
But it is also certainly possible to get the syntax you're describing. You just need a computed variable (which can be a local variable).
let index = 0 // Just to show it can be externally configurable
var referenceToCardStruct: Card {
get { cards[index] }
set { cards[index] = newValue }
}
referenceToCardStruct.name0 = "good"
referenceToCardStruct.name1 = "bye"
print(cards[0].name0 + " " + cards[0].name1)
struct Card {
var name0: String
var name1: String
}
var cards = [Card]()
// every time I want to mutate a struct property :(
cards[0].name0 = "good"
cards[0].name1 = "bye"
Instead of having to perform multiple array index accessing every time I want to mutate a property in struct, is there a technique to avoid such repeating array index accessing operation?
No. When you have an array of struct, then in order to make a change to a struct within the array, you must refer to that struct by index.
If you don't want to see the repeated use of the index, you can hide it in a function using inout:
func mutate(card: inout Card) {
card.name0 = "good"
card.name1 = "bye"
}
for index in cards.indices {
mutate(card:&cards[index])
}
Some day, Swift may include for inout which will allow you to cycle through an array of struct and mutate each struct instance directly. But that day is not yet here.
In answer to the implied question whether it is worth switching to a class just to avoid this repeated use of the index, my answer would be No. There is a good reason for using structs — they are much easier to reason about than classes, and are one of Swift's best features — and I would keep using them if that reason matters to you.
struct is a value type you can't get a reference to it's object with assignment , you should go that way , use a mutating method like https://stackoverflow.com/a/52497495/5820010 or use a class instead
If you don't want to repeat the index, then create a variable from the value you want.
var cards = [Card]()
cards.append(Card(name0: "hello", name1: "world"))
var card = cards[0]
card.name0 = "good"
card.name1 = "bye"
// ...
// ...
cards[0] = card // update the array with the updated card
// "good bye"
print(card.name0 + " " + card.name1)
I think the mutating method is the way to go, as Sh_Khan points out.
In your case, I would do something like:
1> struct Card {
2. var name0: String
3. var name1: String
4. }
5.
6. var cards = [Card]()
7. cards.append(Card(name0: "hello", name1: "world"))
cards: [Card] = 1 value {
[0] = {
name0 = "hello"
name1 = "world"
}
}
8> extension Card {
9. mutating func setNames(name0: String, name1: String) {
10. self.name0 = name0
11. self.name1 = name1
12. }
13. }
14> cards[0].setNames(name0: "x", name1: "y")
15> cards
$R0: [Card] = 1 value {
[0] = {
name0 = "x"
name1 = "y"
}
}
Another approach is a wholesale update of cards[0].
Kind of makes you wish for record updating syntax (a la Haskell or Elm) or dictionary merging-type functionality. But look on the bright side. Maybe Swift's lack of making this easy is testament to the fact that it has static-typing-and-value-semantics-while-allowing-mutation and that combination of features, I think, makes the mutating func or full array element update all but required. I'm not sure Swift has a syntactic feature for updating multiple properties in one shot (without writing your own function or method).
I want to refactor this Swift code with a closure syntax
var station: Station!
var allStations = [Station]()
var favoriteStationIds = [Int]()
for favoriteStationId in favoriteStationIds {
for station in allStations {
if station.stationId == favoriteStationId {
station.isFavorite = true
continue
}
}
}
You can use forEach, which have a trailing closure syntax instead of normal for ... in loops.
Moreover, you don't need to manually iterate through both arrays, you can use index(where:), that accepts a closure to find the station with the specified id as favoriteStationId.
favoriteStationIds.forEach{ id in
allStations[allStations.index(where: {$0.stationId == id})!].isFavorite = true
}
Bear in mind that above piece of code assumes all elements in favoriteStationIds are valid ids that are present in allStations (if this is not the case, use optional binding for the index instead of force unwrapping).
The Swift book does not clearly document the semantics of non-object types in conjunction with subscripts. The most obvious example is a Dictionary whose value type is an Array: Array is a struct. Therefore, when a subscript returns it, we should expect it to be copied. This, however, is very inconvenient. Fortunately, it seems also not to be the case—at least not in Xcode 8.2.1 (Swift 3.1?).
Consider these examples:
var a: [Int] = [0] // [0]
var b = a // [0]
b.append(1)
b // [0, 1]
a // [0]
As we expect, the array a is copied when it is assigned to b. In contrast,
var h: [Int: [Int]] = [0: [0]] // [0: [0]]
h[0]!.append(1)
h[0] // [0, 1]
If the subscript were simply returning the value using ordinary semantics, we would expect that h[0] would still equal [0] after h[0]!.append(1), but in fact it is the array in the dictionary that is appended to. We can even see if we like that the value remains at the same location in memory, suggesting that it is semantically the same array that was appended to. (Being at the same location does not imply that, but it is not in conflict with that, either.)
var h: [Int: [Int]] = [0: [0]] // [0: [0]]
var aptr: UnsafePointer<[Int]>? = nil
withUnsafePointer(to: &h[0]!) { aptr = $0 }
aptr // UnsafePointer(0x7FFF5351C2C0)
h[0]!.append(1)
withUnsafePointer(to: &h[0]!) { aptr = $0 }
aptr // UnsafePointer(0x7FFF5351C2C0)
This fortunate but seemingly-undocumented behavior does not apply only to Arrays as Dictionary values.
struct S: CustomStringConvertible {
var i: Int = 0
var description: String { return "S(i=\(self.i))" }
}
var g: [Int: S] = [0: S()]
g[0]! // S(i=0)
g[0]!.i = 5
g[0]! // S(i=5)
And, in fact, it does not even apply only to the subscripts of Dictionary types.
struct T {
var s: S? = S()
subscript(x: Int) -> S? {
get {
return self.s
}
set(s) {
self.s = s
}
}
}
var t = T()
t[0]! // S(i=0)
var tptr: UnsafePointer<T>? = nil
withUnsafePointer(to: &t) { tptr = $0 }
tptr // UnsafePointer(0x1007F6DB8)
t[0]!.i = 5
t[0]! // S(i=5)
withUnsafePointer(to: &t) { tptr = $0 }
tptr // UnsafePointer(0x1007F6DB8)
Notably, this does somehow involve the setter: If the set under subscript is removed from the definition of T, the statement t[0]!.i = 5 produces the error
error: cannot assign to property: subscript is get-only
My preferred answer would be a pointer to some Swift documentation that clearly explains the semantics of modifications to non-object values obtained through subscripts. It appears that the language behaves as I would like it to, but I'm not comfortable relying on this behavior while it seems inconsistent with the documentation.
Array is implemented using copy-on-write, so it is not in fact copied each time it is assigned but only when it needs to be as determined by its internal state. It seems this behaviour is implemented in such a way that it is not triggered after being returned from a subscript.
update
Array subscripts are implemented using addressors, essentially the subscript accesses the array element directly. You can read some details in this document: https://github.com/apple/swift/blob/master/docs/proposals/Accessors.rst
Particularly note the section Mixed addressors, quote: 'Mixed addressors have now been adopted by Array to great success', perhaps dictionaries are using the same system now.
Also see this tweet: https://mobile.twitter.com/slava_pestov/status/778488750514970624
by someone who apparently works on the swift compiler at Apple, the related thread has some interesting links.
But essentially the answer to your question is that this isn't documented in a user friendly way, it is using special optimisations behind the scenes to make sure it works efficiently. I agree that a detailed explanation in the documentation would be helpful!
If you are concerned at using undocumented behaviour in your code a workaround would be to wrap your value type in a class before storing it in an array or dictionary. But it seems unlikely that the Swift team will make a change that breaks existing code in this way.
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 }