Swift how to "pass by value" of a object - ios

I am quite new in Swift. And I create a class(for example):
class Fraction{
var a: Int
init(a:Int){
self.a = a
}
func toString() -> String{
return "\(self.a)"
}
}
and I also build a in other class function:
class func A_plusplus(f:Fraction){
f.a++
}
Then in the executive class I write:
var object = Fraction(a:10)
print("before run func = " + object.toString())
XXXclass.A_plusplus(object)
print("after ran func =" + object.toString() )
So the console output is
before run func = 10; after ran func =11
The question is how can I just send a copy of the "object" to keep its value which equal to 10
And if functions are always pass-by-reference, why we still need the keyword: "inout"
what does difference between A_plusplus(&object)//[if I make the parameter to be a inout parameter] and A_plusplus(object)
Universally, I don't want to use struct. Although this will solve my
problem exactly, I do pass-by-value rarely.So I don't want program's
copying processes slow my user's phone down :(
And It seems conforming the NSCopying protocol is a good option.But
I don't know how to implement the function:
func copyWithZone(zone:
NSZone)-> AnyObject? correctly

If your class is subclass of NSObject,better to use NSCopying
class Fraction:NSObject,NSCopying{
var a:Int
var b:NSString?
required init(a:Int){
self.a = a
}
func toString() -> String{
return "\(self.a)"
}
func copyWithZone(zone: NSZone) -> AnyObject {
let theCopy=self.dynamicType.init(a: self.a)
theCopy.b = self.b?.copy() as? NSString
return theCopy
}
}
class XXXclass{
class func A_plusplus(f:Fraction){
f.a++
f.b = "after"
}
}
var object = Fraction(a:10)
object.b = "before"
print("before run func = " + object.toString())
print(object.b!) //“Before”
XXXclass.A_plusplus(object.copy() as! Fraction)
print("after ran func =" + object.toString() )
print(object.b!)//“Before”
If it is just a common swift class,You have to create a copy method
class Fraction{
var a: Int
init(a:Int){
self.a = a
}
func toString() -> String{
return "\(self.a)"
}
func copy()->Fraction{
return Fraction(a: self.a)
}
}
class XXXclass{
class func A_plusplus(f:Fraction){
f.a++
}
}
var object = Fraction(a:10)
print("before run func = " + object.toString())
XXXclass.A_plusplus(object.copy())
print("after ran func =" + object.toString() )
To make it clear,you have to know that there are mainly two types in swift
Reference types. Like Class instance,function type
Value types,Like struct and others(Not class instance or function type)
If you pass in a Reference types,you pass in the copy of Reference,it still point to the original object.
If you pass in a Copy type,you pass in the copy of value,so it has nothing to do with the original value
Let us talk about inout,if you use it,it pass in the same object or value.It has effect on Value type
func add(inout input:Int){
input++
}
var a = 10
print(a)//10
add(&a)
print(a)//11

Swift has a new concept so called "struct"
You can define Fraction as struct (Not class)
And
struct Fraction{
...
}
var object = Fraction(a:10)
var object1 = object //then struct in swift is value type, so object1 is copy of object (not reference)
And if you use struct then try to use inout in A_plusplus function
Hope this will help you.

how can I just send a copy of the "object" to keep its value which equal to 10
In Swift classes and functions are always passed by reference. Structs, enums and primitive types are passed by value. See this answer.
You can't pass an object by value. You would have to manually copy it before passing it by reference (if that's what you really want).
Another way is to turn your class into a struct, since it would then be passed by value. However, keep in mind there a few other differences between classes and structs, and it might not necessarily be what you want.
And if functions are always pass-by-reference, why we still need the keyword: "inout"
According to the swift documentation, inout is used when
you want a function to modify a parameter’s value, and you want those changes to persist after the function call has ended, define that parameter as an in-out parameter instead.
So in practice with inout you can pass a value type (such as struct or primitive) by reference. You shouldn't really use this very often. Swift provides tuples, that could be used instead.
what does difference between A_plusplus(&object)//[if I make the parameter to be a inout parameter] and A_plusplus(object)
There is no difference for your A_plusplus function. In that function you don't modify the parameter f itself, you modify the f.a property.
The following example shows the effect of using inout when passing a class object. Both functions are the same, differing only in its parameter definition.
class Person {
var name: String
init(name: String) { self.name = name }
}
var me = Person(name: "Lennon") // Must be var to be passed as inout
// Normal object by reference with a var
func normalCall(var p: Person) {
// We sure are able to update p's properties,
// and they will be reflected back to me
p.name = "McCartney"
// Now p points to a new object different from me,
// changes won't be reflected back to me
p = Person(name: "Ringo")
}
// Inout object reference by value
func inoutCall(inout p: Person) {
// We still can update p's properties,
p.name = "McCartney"
// p is an alias to me, updates made will persist to me
p = Person(name: "Ringo")
}
print("\(me.name)") //--> Lennon
normalCall(me)
print("\(me.name)") //--> McCartney
inoutCall(&me)
print("\(me.name)") //--> Ringo
In normalCall p and me are different variables that happen to point to the same object. When you instantiate and assign a new object to p, they no longer refer to the same object. Hence, further changes to this new object will not be reflected back to me.
Stating that p is a var argument just means that its value can change throughout the function, it does not mean the new value will be assigned to what was passed as argument.
On the other hand, in inoutCall you can think of p and me as aliases. As such, assigning a new object to p is the exact same as assigning a new object to me. Any and every change to p is persisted in me after the function ends.

Related

Using init as a Closure

Recently I saw the following code line in a book (about CoreData)
return modelURLs(in: modelName).compactMap(NSManagedObjectModel.init)
I know what the code does but the question is: Why and how does it work?
There should be a closure as the argument of the compactMap function but there's only a "NSManagedObjectModel.init" in NORMAL parenthesis. What's the secret about it? What is it doing there? I would understand it if there's a static/class property called init which returns a closure but I don't think there is.
Unfortunately the book doesn't say more about this line of code. I would like to have further readings from the apple docs but I can't find anything. When I make a google search about "init in closures" then I don't get helpful results.
So you guys are my last hope :)
By the way: the function modelURLs(in: modelName) returns an Array of URLs but that's not really important here.
When using closures different syntax can be used as in the below example that converts an int array to a string array
let array = [1, 2, 3]
The following calls to compactMap will all correctly convert the array and generate the same result
let out1 = array.compactMap({return String($0)})
let out2 = array.compactMap({String($0)})
let out3 = array.compactMap {String($0)}
let out4 = array.compactMap(String.init)
When there are two init methods that takes the same number and types of argument then you must add the full signature for the init method to use. Consider this simple example struct
struct TwoTimesInt: CustomStringConvertible {
let value: Int
let twiceTheValue: Int
var description: String {
return "\(value) - \(twiceTheValue)"
}
init(value: Int) {
self.value = value
self.twiceTheValue = 2 * value
}
}
With only 1 init method we can do
let out5 = array.compactMap(TwoTimesInt.init)
But if we add a second init method
init(twiceTheValue: Int) {
self.value = twiceTheValue / 2
self.twiceTheValue = twiceTheValue
}
Then we need to give the full signature of the init method to use
let out6 = array.compactMap( TwoTimesInt.init(value:) )
Another thing worth mentioning when it comes to which method is selected is to look at the full signature of the init method including if it returns an optional value or not. So for example if we change the signature of the second init method to return an optional value
init?(twiceTheValue: Int) {
self.value = twiceTheValue / 2
self.twiceTheValue = twiceTheValue
}
then compactMap will favour this init since it expects a closure that returns an optional value, so if we remove the argument name in the call
let out7 = array.compactMap(TwoTimesInt.init)
will use the second init while the map function on the other hand will use the first init method if called the same way.
let out8 = array.map(TwoTimesInt.init)

What's the difference between : and = in swift

Sorry if the title is rather confusing, but I'm curious to know the difference between these two lines:
var title = String()
var title: String
Is one being initialized and one only be declared? Which is more correct?
For example, if I have a struct should I use one of the other?
So the reason I ask this is because I'm learning about how to grab some JSON from a url and then display it in my app. One of the new ways of doing so is using Decodable. So, I have a struct in a model class like so:
struct Videos: Decodable {
var title = String()
var number_of_views : Int
var thumbnail_image_name: String
var channel: Channel
var duration: Int
}
In another class I have this:
URLSession.shared.dataTask(with: url){(data,response,error) in
if(error != nil){
print(error!)
return
}
guard let data = data else { return }
do{
self.Videos2 = try JSONDecoder().decode([Videos].self, from: data)
//self.collectionView?.reloadData()
}catch let jsonErr{
print(jsonErr)
}
}.resume()
So, should I declare or initialize the variables in my struct? I'm assuming I should just declare them like so:
var title: String?
Would that be the correct syntax in my struct?
UPDATE:
I understand this question was more broad then I originally proposed it to be. I'm sorry about that, but thank you so much for all your great answers that clarified a lot up for me.
The difference is that : defines the type of your variable, whereas = assigns an actual value to the variable.
So:
var title = String()
This calls the initializer of the String type, creating a new String instance. It then assigns this value to title. The type of title is inferred to be String because you're assigning an object of type String to it; however, you could also write this line explicitly as:
var title: String = String()
This would mean you are declaring a title variable of type String, and assigning a new String to it.
var title: String
This simply says you're defining a variable of type String. However, you are not assigning a value to it. You will need to assign something to this variable before you use it, or you will get a compile error (and if this is a property rather than just a variable, you'll need to assign it before you get to the end of your type's init() method, unless it's optional with ? after it, in which case it gets implicitly initialized to nil).
EDIT: For your example, I'd probably declare all the variables using let and :, assuming that your JSON provides values for all of those properties. The initializer generated by Decodable should then set all the properties when you create the object. So, something like:
struct Videos: Decodable {
let title: String
let number_of_views : Int
let thumbnail_image_name: String
let channel: Int
let duration: Int
}
This initializes a value
var title = String()
This declares a value but does not initialize it
var title: String
If you attempt to use the latter, such as print(title), you will get a compiler error stating Variable 'title' used before being initialized
It does not matter whether the value is a class or a struct.
The = operator is the assignment operator, it assigns a value to the object on the left of the =
Typically, class or struct properties are declared but not initialized until the init() is called. A simple class might be
class MyClass {
let myProperty: String
init(aString: String) {
self.myProperty = aString
}
}
Whereas inside the scope of a function you may declare a local variable that only lives inside the scope of the function.
func doSomethingToAString(aString: String) -> String {
let extraString = "Something"
let amendedString = aString + extraString
return amendedString
}
In your specific example, the struct synthesizes an initializer that will allow you to initialize the struct with all the values needed to fill your properties. The initializer generated by Decodable should then set all the properties when you create a Videos struct, you will do it something like:
let aVideos = Videos(title: "My Title", number_of_views: 0, thumbnail_image_name: "ImageName", channel: Channel(), duration: 10)
Is one being initialized and one only be declared?
Yes, meaning that the declared cannot be used. If you tried to set a value for it, you would get a compile-time error:
variable 'title' passed by reference before being initialized
Which is more correct?
There is no rule of thumb to determine which is more correct, that would be depends on is there a need to initialize title directly.
On another hand, when it comes to declare properties for a class, saying var title = String() means that you are give title an initial value ("") which means that you are able to create an instance of this class directly, example:
class Foo {
var title = String()
}
let myFoo = Foo()
However, if title declared as var title: String, you will have to implement the init for Foo:
class Foo {
var title: String
init(title: String) {
self.title = title
}
}
let myFoo = Foo(title: "")
Also, you have an option to declare it as lazy:
lazy var title = String()
which means:
A lazy stored property is a property whose initial value is not
calculated until the first time it is used. You indicate a lazy stored
property by writing the lazy modifier before its declaration.
Properties - Lazy Stored Properties

didSet in Swift has a weird knock-on effect on mutating func

I just learned that mutating func is just a curried func with first parameter as inout, so the code below will work and change firstName to "John"
struct Person {
var firstName = "Matt"
mutating func changeName(fn: String) {
firstName = fn
}
}
var p = Person()
let changer = Person.changeName
changer(&p)("John")
p.firstName
but the odd thing happend when I add property observer to p like below, you can see firstName is still "Matt", why?
An interesting note to take ist that the observer is called before the curried setter is called:
struct Person {
var firstName = "Matt"
mutating func changeName(fn: String) {
firstName = fn
}
}
var p: Person = Person() {
didSet {
print("p was set")
}
}
print("1: \(p.firstName)")
let changer = Person.changeName
print("2: \(p.firstName)")
let setter = changer(&p)
print("3: \(p.firstName)")
setter("John")
print("4: \(p.firstName)")
p.changeName("John")
print("5: \(p.firstName)")
This prints:
1: Matt
2: Matt
p was set
3: Matt
4: Matt
p was set
5: John
So it seems that acquiring the setter method on the inout struct performs the actual mutation. This is explained by how inout parameters work semantically: When the parameter is passed to the function its value is copied to a place where the function can mutate it. When the function returns, the value is copied back to the original place, triggering setter observers once, whether the value did change or not.
When we change the way to get the pre-filled curried setter to:
let setter = p.changeName
... the compiler objects:
error: partial application of 'mutating' method is not allowed
It seems that the compiler understands that closing over an inout value is a bad idea, as it is basically taking a reference on a value type.
The closure would let you change the value of the struct at any time, even when the compiler assumes it to be constant. To prevent this unfortunate situation, the compiler simply forbids closing over the inout.
You found a case which fools the compiler and works around the diagnostic. This seems to be an error and it should be filed.
Short version:
struct Foo {
mutating func foo() {}
}
var f = Foo()
let m = Foo.foo
let s = m(&f)
One of the last two lines should emit an error, similar to let x = f.foo.
in the newest accepted proposal 0042-flatten-method-type, self is no more passed as curried function, so this problem is solved in the Swift 3

What should my Array+RemoveObject.Swift contain?

I am using a template project to try to build a new app. When I run the template project, I get an error saying: '[T] does not have a member named 'indexOf'.
The existing code in the Array+RemoveObject.swift doc is:
import Foundation
public func removeObject<T: Equatable>(object: T, inout fromArray array: [T])
{ let index = array.indexOf(object)
if let index = index {
array.removeAtIndex(index)
}
}
Is the problem the use of indexOf? The odd thing is that when I tried using the solution of someone who answered a similar question, I got around 100 errors from the bond framework.
You function on its own works fine for me (Swift 2.1.1, Xcode 7.2). It seems as if you want this function to be a method of a public class of yours. For a minimum working example, you need at least to wrap your removeObject() method within the class you want it to belong to. Note also that you needn't use a separate line for assigning the result from .indexOf(..) call (possibly nil), but can add assignment and nil check in a single if let statement.
public class MyArrayOperations {
// ...
static public func removeObject<T: Equatable>(object: T, inout fromArray array: [T]) {
if let index = array.indexOf(object) {
array.removeAtIndex(index)
}
}
}
var arr = ["1", "2","3"]
MyArrayOperations.removeObject("2", fromArray: &arr)
print(arr) // ["1", "3"]
Also note that you can explicitly state the same behaviour using two generics, in case you want the array itself to conform to some protocol. You then use one separate generic for the array type and one for its elements, thereafter specifying that the element type should conform to the Generator.Element type of the array. E.g.:
func removeObject<T: Equatable, U: _ArrayType where U.Generator.Element == T>(object: T, inout fromArray array: U) {
if let index = array.indexOf(object) {
array.removeAtIndex(index)
}
}
With this approach, you could add an additional protocol at type constraint for the array generic U in the function signature above, e.g.
func removeObject<T: Equatable, U: protocol<_ArrayType, MyProtocol> where U.Generator.Element == T>(object: T, inout fromArray array: [T]) { ...
This can be especially useful when "simulating" generic Array extensions that conforms to some protocol. See e.g. the following for such an example:
extend generic Array<T> to adopt protocol

Changes on Array not being kept after function exits

I've essentially got two methods:
private func addBagToCollection(bag : VendorBag){
var array = arrayForService(bag.type)
array.append(bag)
}
And
func arrayForService(type : BAG_TYPE) -> Array<VendorBag>{
switch type{
case .bagTypeDryCleaning:
return dcBags
case .bagTypeWashAndFold:
return wfBags
case .bagTypeLaunderedShirt:
return lsBags
case .bagTypeHangDry:
return hdBags
case .bagTypeUnknown:
return unBags
}
}
The issue is that the array being referenced in addBagToCollection isn't saving the appended items, meaning that every time that I call the method, the size of the array is 0.
I initialize all my arrays at the top of the class:
var hdBags : Array<VendorBag> = [VendorBag]()
var wfBags : Array<VendorBag> = [VendorBag]()
var lsBags : Array<VendorBag> = [VendorBag]()
var dcBags : Array<VendorBag> = [VendorBag]()
var unBags : Array<VendorBag> = [VendorBag]()
but for some reason, the arrayForService function seems to either only return a copy of the array, or a newly initialized one. How do I return a reference to that array so that all changes made to it are kept even after exiting the function?
Got some more info:
So indeed it seems like arrays are copied when returned from functions--
Swift’s Array types are implemented as structures. This means that arrays
are copied when they are assigned to a new constant or variable, or when they are passed to a function or method.
So how do I return a reference to the actual array rather than a copy of it?
Thanks!
Unlike NSArray class, Swift arrays are struct types which means they are copied when they are passed into functions.
From docs:
Note
Structures are always copied when they are passed around in your code, and do not use reference counting.
You need to use inout keyword if you want your array to be passed by reference, not by copy:
private func addBagToCollection(inout array: Array<VendorBag>, bag: VendorBag) {
array.append(bag)
}
You can call that function as follows:
addBagToCollection(&arrayForService(bag.type), bag)

Resources