Is conditional binding in swift pass by value or reference? - swift

I've been having some issues with conditional binding returning an invalid (but non-nil) object from the watch accelerometer. I was thinking maybe making a copy of the object could help the problem, but I wasn't sure if that was already occurring. If I use code such as:
if let data = recorder.accelerometerData(from: startDate, to: endDate){...}
is this already creating a copy of the CMSensorDataList object or am I simply getting a reference to it?

It just depends upon whether the type wrapped by the optional was a value type or reference type. If reference type, it's obviously pass by reference. If value type, it's copied (unless CoW, copy-on-write, in which case it's copied if and when it's mutated).
In this case, CMSensorDataList is a class, so it's a reference to that instance, not a copy of it.

Related

How am I able to change this constant? [duplicate]

I'm really new to Swift and I just read that classes are passed by reference and arrays/strings etc. are copied.
Is the pass by reference the same way as in Objective-C or Java wherein you actually pass "a" reference or is it proper pass by reference?
Types of Things in Swift
The rule is:
Class instances are reference types (i.e. your reference to a class instance is effectively a pointer)
Functions are reference types
Everything else is a value type; "everything else" simply means instances of structs and instances of enums, because that's all there is in Swift. Arrays and strings are struct instances, for example. You can pass a reference to one of those things (as a function argument) by using inout and taking the address, as newacct has pointed out. But the type is itself a value type.
What Reference Types Mean For You
A reference type object is special in practice because:
Mere assignment or passing to function can yield multiple references to the same object
The object itself is mutable even if the reference to it is a constant (let, either explicit or implied).
A mutation to the object affects that object as seen by all references to it.
Those can be dangers, so keep an eye out. On the other hand, passing a reference type is clearly efficient because only a pointer is copied and passed, which is trivial.
What Value Types Mean For You
Clearly, passing a value type is "safer", and let means what it says: you can't mutate a struct instance or enum instance through a let reference. On the other hand, that safety is achieved by making a separate copy of the value, isn't it? Doesn't that make passing a value type potentially expensive?
Well, yes and no. It isn't as bad as you might think. As Nate Cook has said, passing a value type does not necessarily imply copying, because let (explicit or implied) guarantees immutability so there's no need to copy anything. And even passing into a var reference doesn't mean that things will be copied, only that they can be if necessary (because there's a mutation). The docs specifically advise you not to get your knickers in a twist.
Everything in Swift is passed by "copy" by default, so when you pass a value-type you get a copy of the value, and when you pass a reference type you get a copy of the reference, with all that that implies. (That is, the copy of the reference still points to the same instance as the original reference.)
I use scare quotes around the "copy" above because Swift does a lot of optimization; wherever possible, it doesn't copy until there's a mutation or the possibility of mutation. Since parameters are immutable by default, this means that most of the time no copy actually happens.
It is always pass-by-value when the parameter is not inout.
It is always pass-by-reference if the parameter is inout. However, this is somewhat complicated by the fact you need to explicitly use the & operator on the argument when passing to an inout parameter, so it may not fit the traditional definition of pass-by-reference, where you pass the variable directly.
Here is a small code sample for passing by reference.
Avoid doing this, unless you have a strong reason to.
func ComputeSomeValues(_ value1: inout String, _ value2: inout Int){
value1 = "my great computation 1";
value2 = 123456;
}
Call it like this
var val1: String = "";
var val2: Int = -1;
ComputeSomeValues(&val1, &val2);
The Apple Swift Developer blog has a post called Value and Reference Types that provides a clear and detailed discussion on this very topic.
To quote:
Types in Swift fall into one of two categories: first, “value types”,
where each instance keeps a unique copy of its data, usually defined
as a struct, enum, or tuple. The second, “reference types”, where
instances share a single copy of the data, and the type is usually
defined as a class.
The Swift blog post continues to explain the differences with examples and suggests when you would use one over the other.
When you use inout with an infix operator such as += then the &address symbol can be ignored. I guess the compiler assumes pass by reference?
extension Dictionary {
static func += (left: inout Dictionary, right: Dictionary) {
for (key, value) in right {
left[key] = value
}
}
}
origDictionary += newDictionaryToAdd
And nicely this dictionary 'add' only does one write to the original reference too, so great for locking!
Classes and structures
One of the most important differences between structures and classes is that structures are always copied when they are passed around in your code, but classes are passed by reference.
Closures
If you assign a closure to a property of a class instance, and the closure captures that instance by referring to the instance or its members, you will create a strong reference cycle between the closure and the instance. Swift uses capture lists to break these strong reference cycles
ARC(Automatic Reference Counting)
Reference counting applies only to instances of classes. Structures and enumerations are value types, not reference types, and are not stored and passed by reference.
Classes are passed by references and others are passed by value in default.
You can pass by reference by using the inout keyword.
Swift assign, pass and return a value by reference for reference type and by copy for Value Type
[Value vs Reference type]
If compare with Java you can find matches:
Java Reference type(all objects)
Java primitive type(int, bool...) - Swift extends it using struct
struct is a value type so it's always passed as a value. let create struct
//STEP 1 CREATE PROPERTIES
struct Person{
var raw : String
var name: String
var age: Int
var profession: String
// STEP 2 CREATE FUNCTION
func personInformation(){
print("\(raw)")
print("name : \(name)")
print("age : \(age)")
print("profession : \(profession)")
}
}
//allow equal values
B = A then call the function
A.personInformation()
B.personInformation()
print(B.name)
it have the same result when we change the value of 'B' Only Changes Occured in B Because A Value of A is Copied, like
B.name = "Zainab"
a change occurs in B's name. it is Pass By Value
Pass By Reference
Classes Always Use Pass by reference in which only address of occupied memory is copied, when we change similarly as in struct change the value of B , Both A & B is changed because of reference is copied,.

Class vs. Struct in Swift (copying)

I am trying to understand the concept of why struct vs. class have difference results. Why is the result the same here but different on structs:
import UIKit
class Message {
var internalText: String = "This is some text"
}
// create new instance
var firstMessage = Message()
//if I assign, its a reference to the original instance
var secondMessage = firstMessage
secondMessage.internalText += " with some more text added on."
//print both
print(firstMessage.internalText)
print(secondMessage.internalText)
output:
This is some text with some more text added on.
This is some text with some more text added on.
Now if you change the above from declaration from "class" to "struct"
struct Message {
var internalText: String = "This is some text"
}
...
output becomes:
This is some text
This is some text with some more text added on.
Why in the class declaration does it change the firstMessage object. Are they the same objects? Is this a rule that if I assign a new object from the old object? Then I would have to declare secondMessage = Message() to make it a new instance.
Thanks in advance.
In Swift, classes are reference types, whereas structs are value types. Value types are copied on variable assignment, whereas reference types are not.
More explanation
The system stores instantiated classes and structs into the memory. There are two main sections of the memory involved in the storage of data, the stack, and the heap. The stack contains the local variables introduced in the current method or function, and the heap is used as a kinda external memory, storing larger values. The program can only access variables stored in the stack, so a reference to the value in the heap should be held in the stack.
When you instantiate a class object by using something like Message(), a free space is reserved in your memory's heap and a reference to it is held in the stack. When you assign the same variable to a new one, the reference is copied and both variables will refer to the same bytes in the heap, so changing one changes another too.
When using structs, all the space is being reserved on the stack and there is no such thing as a pointer or reference, so when assigning to a new variable, all the data gets copied (in fact, the system is smart enough to only copy the necessary values which are being changed).
You can see a nice tutorial covering these subjects here.
Why in the class declaration does it change the firstMessage object. Are they the same objects?
The example you gave is a really nice one because it succinctly illustrates the difference between class and struct, and you came about this close -> <- to answering your own question, even if you didn't realize it. As the other answers have explained, class creates a reference type, which means that when you assign an instance of a class to a variable, that variable gets a reference to the object, not a copy of it. You said so yourself:
//if I assign, its a reference to the original instance
var secondMessage = firstMessage
In your example, firstMessage and secondMessage are really references to the one object that you created. This kind of thing is done all the time in object oriented languages because it's often important to know that you're dealing with a specific object and not a copy, especially if you might want to make changes to that object. But that also brings danger: if your code can get a reference to an object and change it, so can some other code in the program. Shared objects that can be changed create all kinds of headaches when you start writing multithreaded code. When you added text to secondMessage, firstMessage also changed because both variables refer to the same object.
Changing the declaration of Message to struct makes it a value type, where assignment (for example) creates a new copy of the object in question instead of a new reference to the same object. When you added text to secondMessage after changing Message to a struct, the assignment secondMessage = firstMessage created a copy of firstMessage, and you only changed that copy.
Is this a rule that if I assign a new object from the old object?
Whether your assignment creates a copy of the object or a reference to it depends, as you've shown, on whether the thing being assigned has reference semantics (class) or value semantics (struct). So you need to be aware of the difference, but most of the time you don't need to think too hard about it. If you're dealing with an object where you don't care about the object's identity and are mainly concerned with its contents (like a number, string, or array), expect that to be a struct. If you care about which object you're dealing with, like the front window or the current document, that'll be a class.
Then I would have to declare secondMessage = Message() to make it a new instance.
Right -- if Message is a class, assigning one to a new variable or passing it into a method won't create a new one. So again, are you more likely to care about which message you're dealing with, or what is in the message?
Simple answer: Classes are reference types Structs are value types.
In the class, firstMessage is set to Message() which is an instance of the whole class Message. So when secondMessage gets set to equal firstMessage, secondMessage Doesn’t make a new class again, it just makes a note of where firstMessage is at and they both can now operate it. But because they both in the same location, the internalText will be the same for both.
While with the struct, since they are value types, secondMessage copies all the values from firstMessage and creates its own independent object of type Message.
Classes are reference types, meaning that the firstMessage and secondMessage variables you defined in your first snippet stores only a reference to the class instance you created. Imagine your object is located somewhere in your memory heap with an id (for example, id0001), then both firstMessage and secondMessage stores only the id, which is id0001, so they both refer to the same object in memory.
On the other hand, structs are value types, meaning that the struct variables store unique objects directly; unlike reference types, no sharing is going on. So when you are assigning a new struct variable to a previous struct variable, the object gets copied, and the two variables store two unique objects with different memory addresses (IDs).
For more information, check out the official doc on classes and structs.
Let us understand the same concept with an example,
Suppose you have a google sheet in which you are adding some text and at a time you share that sheet to some other person for editing or deleting purpose. So when the other person do any changes you can see at a time. This concept is followed in class.
Moreover, classes are reference types because here you are passing a reference(sheet).
However, you have downloaded that google sheet and send its copy to another person so at that time you are not able to see the changes until and unless the person sends back the sheet. And this is the same concept followed in struct. A struct is value type because we are passing a copy(downloaded sheet).
We can inherit class but cannot inherit struct
Think of structs as a Microsoft Excel file. You create a copy and send it to me. When I change my copy, your copy doesn't get changed.
Classes on the other hand are more like Google Sheets. When I make changes to the file you shared with me, you can see the changes.
Instances of structs make copies and have different places in memory
Instances of classes point to the same place in memory

Stopping reference variables changing the value of the original variable

I am assigning the value of a custom class to another variable. Updating the value of the new variable is affecting the value of the original variable. However, I need to stop the reference variable from updating the original variable.
Here's a basic representation of what's happening:
var originalVariable = CustomClass()
originalVariable.myProperty = originalValue
var referenceVariable = originalVariable
referenceVariable.myProperty = updatedValue
print("\(originalVariable.myProperty)") //this prints the ->updatedValue<- and not the ->originalValue<-
I've tried wrapping the referenceVariable in a struct to make it a value type but it hasn't solved the problem.
I've found information regarding value and reference types but I haven't been able to find a solution.
My question in a nutshell: How do I stop an update to a reference variable from updating the original variable that it got its value assigned from?
Thanks in advance.
The whole point of reference semantics (as used by classes) is that all variables point to the same (i.e., they reference the same) object in memory. If you don't want that behaviour, you should use value types (Struct, Enum, Array...) or create copies of your object.
If CustomClass implements the NSCopying protocol you can do:
var referenceVariable = originalVariable.copy()
If it doesn't, you'll have to find some other way to copy it or implement the protocol yourself.
Wrapping the class in a struct will just make two different structs each containing a different reference to the same object.

Why doesn't inout pass by reference?

I'm doing something like this:
someFunction(&myClass)
where someFunction sorts an array on myClass.
someFunction(inout someclass:ClassA) {
someClass.sort({$0.price > $1.price})
}
If I print myClass after the function call, I notice the array is still unsorted. From what I know, Swift passes values by copy. But when I use inout, shouldn't it change to pass by reference?
This is because class instances and functions are reference types. Ints, structs, and everything else are value types. When you pass a reference type into a function as a parameter, you are already going to be referencing that instance. When you pass a value type as a parameter, the function gets a copy of that variable (by default), so inout is usually (see edit) only needed if you want to alter a value type from inside of a function.
Altering a class instance without & or inout:
More details
When you create a reference type var t = myClass(), you're really creating a variable t that is a pointer to a myClass instance in memory. By using an ampersand &t in front of a reference type, you are really saying "give me the pointer to the pointer of a myClass instance"
More info on reference vs value types: https://stackoverflow.com/a/27366050/580487
EDIT
As was pointed out in the comments, you can still use inout with reference types if you want to alter a pointer, etc, but I was trying to shed light on the general use case.
Below is an example of sorting an array inside of a function:
If you post your code here, it would be more meaningful. BTW, look at below links that might helpful for you,
https://developer.apple.com/library/prerelease/ios/documentation/Swift/Conceptual/Swift_Programming_Language/Functions.html#//apple_ref/doc/uid/TP40014097-CH10-ID173
https://developer.apple.com/library/prerelease/ios/documentation/Swift/Conceptual/Swift_Programming_Language/Declarations.html#//apple_ref/doc/uid/TP40014097-CH34-ID545

Is Swift Pass By Value or Pass By Reference

I'm really new to Swift and I just read that classes are passed by reference and arrays/strings etc. are copied.
Is the pass by reference the same way as in Objective-C or Java wherein you actually pass "a" reference or is it proper pass by reference?
Types of Things in Swift
The rule is:
Class instances are reference types (i.e. your reference to a class instance is effectively a pointer)
Functions are reference types
Everything else is a value type; "everything else" simply means instances of structs and instances of enums, because that's all there is in Swift. Arrays and strings are struct instances, for example. You can pass a reference to one of those things (as a function argument) by using inout and taking the address, as newacct has pointed out. But the type is itself a value type.
What Reference Types Mean For You
A reference type object is special in practice because:
Mere assignment or passing to function can yield multiple references to the same object
The object itself is mutable even if the reference to it is a constant (let, either explicit or implied).
A mutation to the object affects that object as seen by all references to it.
Those can be dangers, so keep an eye out. On the other hand, passing a reference type is clearly efficient because only a pointer is copied and passed, which is trivial.
What Value Types Mean For You
Clearly, passing a value type is "safer", and let means what it says: you can't mutate a struct instance or enum instance through a let reference. On the other hand, that safety is achieved by making a separate copy of the value, isn't it? Doesn't that make passing a value type potentially expensive?
Well, yes and no. It isn't as bad as you might think. As Nate Cook has said, passing a value type does not necessarily imply copying, because let (explicit or implied) guarantees immutability so there's no need to copy anything. And even passing into a var reference doesn't mean that things will be copied, only that they can be if necessary (because there's a mutation). The docs specifically advise you not to get your knickers in a twist.
Everything in Swift is passed by "copy" by default, so when you pass a value-type you get a copy of the value, and when you pass a reference type you get a copy of the reference, with all that that implies. (That is, the copy of the reference still points to the same instance as the original reference.)
I use scare quotes around the "copy" above because Swift does a lot of optimization; wherever possible, it doesn't copy until there's a mutation or the possibility of mutation. Since parameters are immutable by default, this means that most of the time no copy actually happens.
It is always pass-by-value when the parameter is not inout.
It is always pass-by-reference if the parameter is inout. However, this is somewhat complicated by the fact you need to explicitly use the & operator on the argument when passing to an inout parameter, so it may not fit the traditional definition of pass-by-reference, where you pass the variable directly.
Here is a small code sample for passing by reference.
Avoid doing this, unless you have a strong reason to.
func ComputeSomeValues(_ value1: inout String, _ value2: inout Int){
value1 = "my great computation 1";
value2 = 123456;
}
Call it like this
var val1: String = "";
var val2: Int = -1;
ComputeSomeValues(&val1, &val2);
The Apple Swift Developer blog has a post called Value and Reference Types that provides a clear and detailed discussion on this very topic.
To quote:
Types in Swift fall into one of two categories: first, “value types”,
where each instance keeps a unique copy of its data, usually defined
as a struct, enum, or tuple. The second, “reference types”, where
instances share a single copy of the data, and the type is usually
defined as a class.
The Swift blog post continues to explain the differences with examples and suggests when you would use one over the other.
When you use inout with an infix operator such as += then the &address symbol can be ignored. I guess the compiler assumes pass by reference?
extension Dictionary {
static func += (left: inout Dictionary, right: Dictionary) {
for (key, value) in right {
left[key] = value
}
}
}
origDictionary += newDictionaryToAdd
And nicely this dictionary 'add' only does one write to the original reference too, so great for locking!
Classes and structures
One of the most important differences between structures and classes is that structures are always copied when they are passed around in your code, but classes are passed by reference.
Closures
If you assign a closure to a property of a class instance, and the closure captures that instance by referring to the instance or its members, you will create a strong reference cycle between the closure and the instance. Swift uses capture lists to break these strong reference cycles
ARC(Automatic Reference Counting)
Reference counting applies only to instances of classes. Structures and enumerations are value types, not reference types, and are not stored and passed by reference.
Classes are passed by references and others are passed by value in default.
You can pass by reference by using the inout keyword.
Swift assign, pass and return a value by reference for reference type and by copy for Value Type
[Value vs Reference type]
If compare with Java you can find matches:
Java Reference type(all objects)
Java primitive type(int, bool...) - Swift extends it using struct
struct is a value type so it's always passed as a value. let create struct
//STEP 1 CREATE PROPERTIES
struct Person{
var raw : String
var name: String
var age: Int
var profession: String
// STEP 2 CREATE FUNCTION
func personInformation(){
print("\(raw)")
print("name : \(name)")
print("age : \(age)")
print("profession : \(profession)")
}
}
//allow equal values
B = A then call the function
A.personInformation()
B.personInformation()
print(B.name)
it have the same result when we change the value of 'B' Only Changes Occured in B Because A Value of A is Copied, like
B.name = "Zainab"
a change occurs in B's name. it is Pass By Value
Pass By Reference
Classes Always Use Pass by reference in which only address of occupied memory is copied, when we change similarly as in struct change the value of B , Both A & B is changed because of reference is copied,.