Swift: confused about nullable/optional types - swift

I'm new to Swift and iOS coding and have been working on writing my first app. While my programming background is pretty significant, I come from a Python and C# background where pretty much anything can be None or null and it's up to the user to check at runtime for a null. I'm finding this whole concept of "nullable vs. non-nullable types" or "optional types" in Swift to be confusing.
I understand that the core concept is that a variable declared as a type like myObject cannot be set to nil. However, if I define it as type myObject? then the value can be set to nil.
The problem is that, when I look at my code designs, it feels like everything will have to be "nullable" in my code. It feels like this either means I'm not thinking correctly with how my code should run, or that I'm missing some crucial piece of understanding.
Let's take the simplest example of something I am confused about. Suppose I have two classes - one that stores and manages some sort of data, and another that provides access to that data. (An example of this might be something like a database connection, or a file handle, or something similar.) Let's call the class containing data myData and the class that works with that data myObject.
myObject will need a class-level reference to myData because many of its methods depend on a local reference to the class. So, the first thing the constructor does is to generate a data connection and then store it in the local variable dataConnection. The variable needs to be defined at the class level so other methods can access it, but it will be assigned to in the constructor. Failure to obtain the connection will result in some sort of exception that will interfere with the very creation of the class.
I know that Swift has two ways to define a variable: var and let, with let being analogous to some languages' const directive. Since the data connection will persist throughout the entire life of the class, let seems an obvious choice. However, I do not know how to define a class-level variable via let which will be assigned at runtime. Therefore, I use something like
var dataConnection: myData?
in the class outside any functions.
But now I have to deal with the nullable data type, and do explicit unwrapping every time I use it anywhere. It is frustrating to say the least and quite confusing.
func dealWithData() {
self.dataConnection.someFunctionToGetData() <- results in an unwrapping error.
self.dataConnection!.someFunctionToGetData() <- works.
let someOtherObjectUsingData: otherObject = self.getOtherObject() <- may result in error unless type includes ?
someOtherObjectUsingData.someMethod(self.dataConnection) <- unwrap error if type included ?
var myData = self.dataConnection!
someOtherObjectUsingData.someMethod(myData) <- works
}
func somethingNeedingDataObject(dataObject: myData?) {
// now have to explicitly unwrap
let myDataUnwrapped = myData!
...
}
This just seems to be an extremely verbose way to deal with the issue. If an object is nil, won't the explicit unwrap in and of itself cause a runtime error (which could be caught and handled)? This tends to be a nightmare when stringing things together. I've had to do something like:
self.dataConnection!.somethingReturningAnObject!.thatObjectsVariable!.someMethod()
var myData? = self.dataConnection
var anotherObject? = myData!.somethingReturningAnObject
...
The way I'm used to doing this is that you simply define a variable, and if it is set to null and you try to do something with it, an exception (that you can catch and handle) is thrown. Is this simply not the way things work anymore in Swift? This has confused me sufficiently that just about every time I try to compile an app, I get tons of errors about this (and I just let Xcode fix them). But this can't be the best way to deal with it.
Do I have to consistently deal with wrapping and unwrapping variables - even those which are expected to never be null in the first place but simply can't be assigned at compile time?

However, I do not know how to define a class-level variable via let which will be assigned at runtime.
This part is easy. Just use let instead of var. With Swift 1.2 and later, you can delay the actual assignment of a let. The compiler is smart enough to do flow analysis and make sure it's assigned once, and only once, in all paths. So in the case of a class-wide let, the assignment can also happen in the constructor.
But now I have to deal with the nullable data type, and do explicit unwrapping every time I use it anywhere.
But this is what implicitly unwrapped Optionals are for. For example, StoryBoard defines all #IBOutlets as implicitly unwrapped, because the semantics are very clear: upon entrance to viewDidLoad() and everywhere after, unwrapping is safe. If you can prove clear semantics to yourself, you can do the same.
So you have roughly 4 choices:
A) declare at class level as implicitly unwrapped:
let dataConnection: MyData!
And be forced to initialize it in the constructor:
init() {
let whateverObj = someInitialCalculation()
dataConnection = whateverObj.someWayOfGettingTheConnection()
}
And from then on you don't need the '!'; it should be clear that implicit unwrap is always safe.
B) Initialize it right in its declaration if its initialization is reliable and sensible at that point, allowing you to forgo the entire concept of Optionals:
let dataConnection = SomeClass.someStaticMethod()
C) Declare at class level as a var, as implicit optional:
var dataConnection: MyData!
You won't have to init it in the constructor; let it be nil until its value can/should be computed. You still need some flow analysis to prove after a certain point, as in the case of #IBOutlets, accessing it will always be valid
D) The most 'unpredictable' case. Declare it as an explicit optional, because throughout the lifecycle of the class, the data connection will come and go:
var dataConnection: MyData?
func someMethodThatHandlesData() {
if let dC = dataConnection {
dc.handleSomeData()
}
else {
alert("Sorry, no data connection at the moment. Try again later.")
}
}
I think you're imagining that Swift always forces you down path D).
As far as your spaghetti-string code, you want to look into Optional Chaining, and only need to check the end result for nil.

Related

Does passing struct into a function also using copy-on-write for optimisation, or actual copy does happen immediately? [duplicate]

In Swift, when you pass a value type, say an Array to a function. A copy of the array is made for the function to use.
However the documentation at https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/ClassesAndStructures.html#//apple_ref/doc/uid/TP40014097-CH13-XID_134 also says:
The description above refers to the “copying” of strings, arrays, and
dictionaries. The behavior you see in your code will always be as if a
copy took place. However, Swift only performs an actual copy behind
the scenes when it is absolutely necessary to do so. Swift manages all
value copying to ensure optimal performance, and you should not avoid
assignment to try to preempt this optimization.
So does it mean that the copying actually only takes placed when the passed value type is modified?
Is there a way to demonstrate that this is actually the underlying behavior?
Why this is important? If I create a large immutable array and want to pass it in from function to function, I certainly do not want to keep making copies of it. Should I just use NSArrray in this case or would the Swift Array work fine as long as I do not try to manipulate the passed in Array?
Now as long as I do not explicitly make the variables in the function editable by using var or inout, then the function can not modify the array anyway. So does it still make a copy? Granted that another thread can modify the original array elsewhere (only if it is mutable), making a copy at the moment the function is called necessary (but only if the array passed in is mutable). So if the original array is immutable and the function is not using var or inout, there is no point in Swift creating a copy. Right? So what does Apple mean by the phrase above?
TL;DR:
So does it mean that the copying actually only takes placed when the passed value type is modified?
Yes!
Is there a way to demonstrate that this is actually the underlying behavior?
See the first example in the section on the copy-on-write optimization.
Should I just use NSArrray in this case or would the Swift Array work fine
as long as I do not try to manipulate the passed in Array?
If you pass your array as inout, then you'll have a pass-by-reference semantics,
hence obviously avoiding unnecessary copies.
If you pass your array as a normal parameter,
then the copy-on-write optimization will kick in and you shouldn't notice any performance drop
while still benefiting from more type safety that what you'd get with a NSArray.
Now as long as I do not explicitly make the variables in the function editable
by using var or inout, then the function can not modify the array anyway.
So does it still make a copy?
You will get a "copy", in the abstract sense.
In reality, the underlying storage will be shared, thanks to the copy-on-write mechanism,
hence avoiding unnecessary copies.
If the original array is immutable and the function is not using var or inout,
there is no point in Swift creating a copy. Right?
Exactly, hence the copy-on-write mechanism.
So what does Apple mean by the phrase above?
Essentially, Apple means that you shouldn't worry about the "cost" of copying value types,
as Swift optimizes it for you behind the scene.
Instead, you should just think about the semantics of value types,
which is that get a copy as soon as you assign or use them as parameters.
What's actually generated by Swift's compiler is the Swift's compiler business.
Value types semantics
Swift does indeed treat arrays as value types (as opposed to reference types),
along with structures, enumerations and most other built-in types
(i.e. those that are part of the standard library and not Foundation).
At the memory level, these types are actually immutable plain old data objects (POD),
which enables interesting optimizations.
Indeed, they are typically allocated on the stack rather than the heap [1],
(https://en.wikipedia.org/wiki/Stack-based_memory_allocation).
This allows the CPU to very efficiently manage them,
and to automatically deallocate their memory as soon as the function exits [2],
without the need for any garbage collection strategy.
Values are copied whenever assigned or passed as a function.
This semantics has various advantages,
such as avoiding the creation of unintended aliases,
but also as making it easier for the compiler to guarantee the lifetime of values
stored in a another object or captured by a closure.
We can think about how hard it can be to manage good old C pointers to understand why.
One may think it's an ill-conceived strategy,
as it involves copying every single time a variable is assigned or a function is called.
But as counterintuitive it may be,
copying small types is usually quite cheap if not cheaper than passing a reference.
After all, a pointer is usually the same size as an integer...
Concerns are however legitimate for large collections (i.e. arrays, sets and dictionaries),
and very large structures to a lesser extent [3].
But the compiler has has a trick to handle these, namely copy-on-write (see later).
What about mutating
Structures can define mutating methods,
which are allowed to mutate the fields of the structure.
This doesn't contradict the fact that value types are nothing more than immutable PODs,
as in fact calling a mutating method is merely a huge syntactic sugar
for reassigning a variable to a brand new value that's identical to the previous ones,
except for the fields that were mutated.
The following example illustrates this semantical equivalence:
struct S {
var foo: Int
var bar: Int
mutating func modify() {
foo = bar
}
}
var s1 = S(foo: 0, bar: 10)
s1.modify()
// The two lines above do the same as the two lines below:
var s2 = S(foo: 0, bar: 10)
s2 = S(foo: s2.bar, bar: s2.bar)
Reference types semantics
Unlike value types, reference types are essentially pointers to the heap at the memory level.
Their semantics is closer to what we would get in reference-based languages,
such as Java, Python or Javascript.
This means they do not get copied when assigned or passed to a function, their address is.
Because the CPU is no longer able to manage the memory of these objects automatically,
Swift uses a reference counter to handle garbage collection behind the scenes
(https://en.wikipedia.org/wiki/Reference_counting).
Such semantics has the obvious advantage to avoid copies,
as everything is assigned or passed by reference.
The drawback is the danger of unintended aliases,
as in almost any other reference-based language.
What about inout
An inout parameter is nothing more than a read-write pointer to the expected type.
In the case of value types, it means the function won't get a copy of the value,
but a pointer to such values,
so mutations inside the function will affect the value parameter (hence the inout keyword).
In other terms, this gives value types parameters a reference semantics in the context of the function:
func f(x: inout [Int]) {
x.append(12)
}
var a = [0]
f(x: &a)
// Prints '[0, 12]'
print(a)
In the case of reference types, it will make the reference itself mutable,
pretty much as if the passed argument was a the address of the address of the object:
func f(x: inout NSArray) {
x = [12]
}
var a: NSArray = [0]
f(x: &a)
// Prints '(12)'
print(a)
Copy-on-write
Copy-on-write (https://en.wikipedia.org/wiki/Copy-on-write) is an optimization technique that
can avoid unnecessary copies of mutable variables,
which is implemented on all Swift's built-in collections (i.e. array, sets and dictionaries).
When you assign an array (or pass it to a function),
Swift doesn't make a copy of the said array and actually uses a reference instead.
The copy will take place as soon as the your second array is mutated.
This behavior can be demonstrated with the following snippet (Swift 4.1):
let array1 = [1, 2, 3]
var array2 = array1
// Will print the same address twice.
array1.withUnsafeBytes { print($0.baseAddress!) }
array2.withUnsafeBytes { print($0.baseAddress!) }
array2[0] = 1
// Will print a different address.
array2.withUnsafeBytes { print($0.baseAddress!) }
Indeed, array2 doesn't get a copy of array1 immediately,
as shown by the fact it points to the same address.
Instead, the copy is triggered by the mutation of array2.
This optimization also happens deeper in the structure,
meaning that if for instance your collection is made of other collections,
the latter will also benefit from the copy-on-write mechanism,
as demonstrated by the following snippet (Swift 4.1):
var array1 = [[1, 2], [3, 4]]
var array2 = array1
// Will print the same address twice.
array1[1].withUnsafeBytes { print($0.baseAddress!) }
array2[1].withUnsafeBytes { print($0.baseAddress!) }
array2[0] = []
// Will print the same address as before.
array2[1].withUnsafeBytes { print($0.baseAddress!) }
Replicating copy-on-write
It is in fact rather easy to implement the copy-on-write mechanism in Swift,
as some of the its reference counter API is exposed to the user.
The trick consists of wrapping a reference (e.g. a class instance) within a structure,
and to check whether that reference is uniquely referenced before mutating it.
When that's the case, the wrapped value can be safely mutated,
otherwise it should be copied:
final class Wrapped<T> {
init(value: T) { self.value = value }
var value: T
}
struct CopyOnWrite<T> {
init(value: T) { self.wrapped = Wrapped(value: value) }
var wrapped: Wrapped<T>
var value: T {
get { return wrapped.value }
set {
if isKnownUniquelyReferenced(&wrapped) {
wrapped.value = newValue
} else {
wrapped = Wrapped(value: newValue)
}
}
}
}
var a = CopyOnWrite(value: SomeLargeObject())
// This line doesn't copy anything.
var b = a
However, there is an import caveat here!
Reading the documentation for isKnownUniquelyReferenced we get this warning:
If the instance passed as object is being accessed by multiple threads simultaneously,
this function may still return true.
Therefore, you must only call this function from mutating methods
with appropriate thread synchronization.
This means the implementation presented above isn't thread safe,
as we may encounter situations where it'd wrongly assumes the wrapped object can be safely mutated,
while in fact such mutation would break invariant in another thread.
Yet this doesn't mean Swift's copy-on-write is inherently flawed in multithreaded programs.
The key is to understand what "accessed by multiple threads simultaneously" really means.
In our example, this would happen if the same instance of CopyOnWrite was shared across multiple threads,
for instance as part of a shared global variable.
The wrapped object would then have a thread safe copy-on-write semantics,
but the instance holding it would be subject to data race.
The reason is that Swift must establish unique ownership
to properly evaluate isKnownUniquelyReferenced [4],
which it can't do if the owner of the instance is itself shared across multiple threads.
Value types and multithreading
It is Swift's intention to alleviate the burden of the programmer
when dealing with multithreaded environments, as stated on Apple's blog
(https://developer.apple.com/swift/blog/?id=10):
One of the primary reasons to choose value types over reference types
is the ability to more easily reason about your code.
If you always get a unique, copied instance,
you can trust that no other part of your app is changing the data under the covers.
This is especially helpful in multi-threaded environments
where a different thread could alter your data out from under you.
This can create nasty bugs that are extremely hard to debug.
Ultimately, the copy-on-write mechanism is a resource management optimization that,
like any other optimization technique,
one shouldn't think about when writing code [5].
Instead, one should think in more abstract terms
and consider values to be effectively copied when assigned or passed as arguments.
[1]
This holds only for values used as local variables.
Values used as fields of a reference type (e.g. a class) are also stored in the heap.
[2]
One could get confirmation of that by checking the LLVM byte code that's produced
when dealing with value types rather than reference types,
but the Swift compiler being very eager to perform constant propagation,
building a minimal example is a bit tricky.
[3]
Swift doesn't allow structures to reference themselves,
as the compiler would be unable to compute the size of such type statically.
Therefore, it is not very realistic to think of a structure that is so large
that copying it would become a legitimate concern.
[4]
This is, by the way, the reason why isKnownUniquelyReferenced accepts an inout parameter,
as it's currently Swift's way to establish ownership.
[5]
Although passing copies of value-type instances should be safe,
there's a open issue that suggests some problems with the current implementation
(https://bugs.swift.org/browse/SR-6543).
I don't know if that's the same for every value type in Swift, but for Arrays I'm pretty sure it's a copy-on-write, so it doesn't copy it unless you modify it, and as you said if you pass it around as a constant you don't run that risk anyway.
p.s. In Swift 1.2 there are new APIs you can use to implement copy-on-write on your own value-types too

Creating a factory method returning generic

I have an abstract class defined this way:
class BaseCoordinator<ResultType>
Other classes inherit from this one eg.
final class AppCoordinator: BaseCoordinator<Void>
final class AuthFlowCoordinator: BaseCoordinator<AuthFlowCoordinationResult>
final class MainFlowCoordinator: BaseCoordinator<Void>
Now I want to create a factory method. I guess it's sugnature should look like:
func makeCoordinator<T>() -> BaseCoordinator<T>
But, of course, I get error like this one:
Cannot convert return expression of type 'AppCoordinator' to return type 'BaseCoordinator<T>'
What Xcode suggests me is to add as! BaseCoordinator<T>, but I hate this force downcasting (and returning ? isn't satisfying me either, as I'm 100% sure I'll have a correct default object), as I want assurance that at least a default Coordinator will be returned.
It feels like I'm missing something, but I really have no clue what is it. Is it actually possible to make such factory method, or is Swift generics limited?
T must have a predefined type at compile time. Therefore you cannot choose T using the implementation of makeCoordinator() by returning different coordinators with different ResultTypes.
In this case, the caller can choose which value he wants to assign to T, potentially breaking the function. For example the following two calls would be perfectly valid:
let coordinator: BaseCoordinator<Void> = makeCoordinator()
let coordinator: BaseCoordinator<[Int: [String]]> = makeCoordinator()
It would not make sense to use [Int: [String]] as a generic type but it is still possible. Depending on the generic type you choose at the function call, the cast may work or not, which is why a force cast will likely lead to a crash.
An optional cast, as suggested by Tom E would fix the potential crash but it would still not resolve this problem.
Therefore you cannot use a factory pattern for this without erasing ResultType to Any using a wrapper type, which would defeat the purpose of generics.
If you want type safety, you have to create a factory method for each subclass of BaseCoordinator you want to instantiate or just call their initializers manually.
You might want to try as? instead of as!. The former yields an optional, i. e. the result will be nil if the cast doesn’t succeed. This way you have a safe cast without the need of forcing anything.

Variable 'xxx' was never mutated, consider changing to 'let'

Updated to xcode7-beta I run across a new kind of warning. Here is my code
override func layoutAttributesForElementsInRect(rect: CGRect) -> [UICollectionViewLayoutAttributes]? {
var attributes: [UICollectionViewLayoutAttributes]? = super.layoutAttributesForElementsInRect(rect)
if let layoutInfo = self.layoutInfo {
attributes?.append(layoutInfo)
}
return attributes
}
the warning message is
Variable 'attributes' was never mutated, consider changing to 'let' constant
Why does xcode say Variable 'attributes' was never mutated?
Question Update
the warning is gone when I change my code to this
override func layoutAttributesForElementsInRect(rect: CGRect) -> [UICollectionViewLayoutAttributes]? {
var attributes: [UICollectionViewLayoutAttributes]? = super.layoutAttributesForElementsInRect(rect)
if let layoutInfo = self.layoutInfo {
attributes!.append(layoutInfo)
}
return attributes
}
so forced unwrapping can take it away. But it might not be a good thing right?
They talked about this in the WWDC videos and the release notes.
It has always been the case that you get much, much better performance (faster speed, smaller space) if you use let instead of var whenever you can. This tells the compiler that this thing is a constant, not a variable, and that fact allows the compiler to optimize all kinds of things away.
But the compiler can't do that unless you do use let whenever you can. It won't change a var to a let for you.
Therefore, in Swift 2, the compiler does a smarter analysis at build time and warns you if you are using var where you could have used let. Eventually this feature will work properly, at which point you should take the compiler's advice!
This has been driving me crazy too. Whenever I have a class and I change member properties, it still tells me that I should make the variable a constant. For instance:
class X {
var name = ""
}
var xInstance = X()
xInstance.name = "my name"
Near as I can figure, the compiler is correct. xInstance is a constant. It is assigned the pointer to the X instance during its initialization and is never assigned another value. So xInstance is constant. The class that it points to is not constant.
This is more obvious in a language like C or C++ but when you realize that all class instances are really pointers to heap backed classes, this makes sense.
For those who understand C++
xInstance is equivalent to:
X *xInstance
making it a let instead means that it changes to:
X * const xInstance
I think most people would intuitively think that the swift declaration would be equivalent to
X const * xInstance
By declaring a constant with let, you ensure that it can never be changed. This is good for making sure that you don't accidentally change it later, and it (in theory) can help the optimizer generate faster code.
If you declare a variable with var, and you don't intend to change it or call mutating methods on it, using let instead helps you enforce that contract.
You have created that object as var object but value of that object not changing then after better to make it let. That's it.
As per Apple Developer Guidelines, create var object if value of that object is going to change else create let variable. Best practise
Your code implies that attributes could be mutated if self.layoutInfo is non nil
Perhaps the warning is saying that no path leads to self.layoutInfo being non nil and hence attributes has no need of being a var.
Examine what conditions if any can lead to self.layoutInfo having data
Why this warning?
Because in Swift "let" means "value is constant" and allows compiler to do some optimizations, hence "let" is faster than "var",
And since "Swift 2", the compiler warns if we use "var" wherever it detects that we could use "let" instead.
Class vs Struct
However, in Swift the "class" is a reference-type (just like in Java), meaning the "value" is just "a reference and/or pointer to actual data".
Even if we use "let" with "class", the "data" pointed is never constant (and is always mutable).
So, if changing the "data" is a big thing and should not happen unnoticed, consider using struct.
In Swift "struct" is a value-type, and unless it's defined with "var" (or "inout" in "func" parameters), any mutation to the "data" it holds should cause compile error.
Basically same as const keyword available in C and C++ languages, but in C++ we don't need to migrate from "class" to a complete different type, jsut to have mutability sensitive class.
That's a plus for C++, but it fades to nothing, once we see the other C++ overcomplications.

When does the copying take place for swift value types

In Swift, when you pass a value type, say an Array to a function. A copy of the array is made for the function to use.
However the documentation at https://developer.apple.com/library/ios/documentation/Swift/Conceptual/Swift_Programming_Language/ClassesAndStructures.html#//apple_ref/doc/uid/TP40014097-CH13-XID_134 also says:
The description above refers to the “copying” of strings, arrays, and
dictionaries. The behavior you see in your code will always be as if a
copy took place. However, Swift only performs an actual copy behind
the scenes when it is absolutely necessary to do so. Swift manages all
value copying to ensure optimal performance, and you should not avoid
assignment to try to preempt this optimization.
So does it mean that the copying actually only takes placed when the passed value type is modified?
Is there a way to demonstrate that this is actually the underlying behavior?
Why this is important? If I create a large immutable array and want to pass it in from function to function, I certainly do not want to keep making copies of it. Should I just use NSArrray in this case or would the Swift Array work fine as long as I do not try to manipulate the passed in Array?
Now as long as I do not explicitly make the variables in the function editable by using var or inout, then the function can not modify the array anyway. So does it still make a copy? Granted that another thread can modify the original array elsewhere (only if it is mutable), making a copy at the moment the function is called necessary (but only if the array passed in is mutable). So if the original array is immutable and the function is not using var or inout, there is no point in Swift creating a copy. Right? So what does Apple mean by the phrase above?
TL;DR:
So does it mean that the copying actually only takes placed when the passed value type is modified?
Yes!
Is there a way to demonstrate that this is actually the underlying behavior?
See the first example in the section on the copy-on-write optimization.
Should I just use NSArrray in this case or would the Swift Array work fine
as long as I do not try to manipulate the passed in Array?
If you pass your array as inout, then you'll have a pass-by-reference semantics,
hence obviously avoiding unnecessary copies.
If you pass your array as a normal parameter,
then the copy-on-write optimization will kick in and you shouldn't notice any performance drop
while still benefiting from more type safety that what you'd get with a NSArray.
Now as long as I do not explicitly make the variables in the function editable
by using var or inout, then the function can not modify the array anyway.
So does it still make a copy?
You will get a "copy", in the abstract sense.
In reality, the underlying storage will be shared, thanks to the copy-on-write mechanism,
hence avoiding unnecessary copies.
If the original array is immutable and the function is not using var or inout,
there is no point in Swift creating a copy. Right?
Exactly, hence the copy-on-write mechanism.
So what does Apple mean by the phrase above?
Essentially, Apple means that you shouldn't worry about the "cost" of copying value types,
as Swift optimizes it for you behind the scene.
Instead, you should just think about the semantics of value types,
which is that get a copy as soon as you assign or use them as parameters.
What's actually generated by Swift's compiler is the Swift's compiler business.
Value types semantics
Swift does indeed treat arrays as value types (as opposed to reference types),
along with structures, enumerations and most other built-in types
(i.e. those that are part of the standard library and not Foundation).
At the memory level, these types are actually immutable plain old data objects (POD),
which enables interesting optimizations.
Indeed, they are typically allocated on the stack rather than the heap [1],
(https://en.wikipedia.org/wiki/Stack-based_memory_allocation).
This allows the CPU to very efficiently manage them,
and to automatically deallocate their memory as soon as the function exits [2],
without the need for any garbage collection strategy.
Values are copied whenever assigned or passed as a function.
This semantics has various advantages,
such as avoiding the creation of unintended aliases,
but also as making it easier for the compiler to guarantee the lifetime of values
stored in a another object or captured by a closure.
We can think about how hard it can be to manage good old C pointers to understand why.
One may think it's an ill-conceived strategy,
as it involves copying every single time a variable is assigned or a function is called.
But as counterintuitive it may be,
copying small types is usually quite cheap if not cheaper than passing a reference.
After all, a pointer is usually the same size as an integer...
Concerns are however legitimate for large collections (i.e. arrays, sets and dictionaries),
and very large structures to a lesser extent [3].
But the compiler has has a trick to handle these, namely copy-on-write (see later).
What about mutating
Structures can define mutating methods,
which are allowed to mutate the fields of the structure.
This doesn't contradict the fact that value types are nothing more than immutable PODs,
as in fact calling a mutating method is merely a huge syntactic sugar
for reassigning a variable to a brand new value that's identical to the previous ones,
except for the fields that were mutated.
The following example illustrates this semantical equivalence:
struct S {
var foo: Int
var bar: Int
mutating func modify() {
foo = bar
}
}
var s1 = S(foo: 0, bar: 10)
s1.modify()
// The two lines above do the same as the two lines below:
var s2 = S(foo: 0, bar: 10)
s2 = S(foo: s2.bar, bar: s2.bar)
Reference types semantics
Unlike value types, reference types are essentially pointers to the heap at the memory level.
Their semantics is closer to what we would get in reference-based languages,
such as Java, Python or Javascript.
This means they do not get copied when assigned or passed to a function, their address is.
Because the CPU is no longer able to manage the memory of these objects automatically,
Swift uses a reference counter to handle garbage collection behind the scenes
(https://en.wikipedia.org/wiki/Reference_counting).
Such semantics has the obvious advantage to avoid copies,
as everything is assigned or passed by reference.
The drawback is the danger of unintended aliases,
as in almost any other reference-based language.
What about inout
An inout parameter is nothing more than a read-write pointer to the expected type.
In the case of value types, it means the function won't get a copy of the value,
but a pointer to such values,
so mutations inside the function will affect the value parameter (hence the inout keyword).
In other terms, this gives value types parameters a reference semantics in the context of the function:
func f(x: inout [Int]) {
x.append(12)
}
var a = [0]
f(x: &a)
// Prints '[0, 12]'
print(a)
In the case of reference types, it will make the reference itself mutable,
pretty much as if the passed argument was a the address of the address of the object:
func f(x: inout NSArray) {
x = [12]
}
var a: NSArray = [0]
f(x: &a)
// Prints '(12)'
print(a)
Copy-on-write
Copy-on-write (https://en.wikipedia.org/wiki/Copy-on-write) is an optimization technique that
can avoid unnecessary copies of mutable variables,
which is implemented on all Swift's built-in collections (i.e. array, sets and dictionaries).
When you assign an array (or pass it to a function),
Swift doesn't make a copy of the said array and actually uses a reference instead.
The copy will take place as soon as the your second array is mutated.
This behavior can be demonstrated with the following snippet (Swift 4.1):
let array1 = [1, 2, 3]
var array2 = array1
// Will print the same address twice.
array1.withUnsafeBytes { print($0.baseAddress!) }
array2.withUnsafeBytes { print($0.baseAddress!) }
array2[0] = 1
// Will print a different address.
array2.withUnsafeBytes { print($0.baseAddress!) }
Indeed, array2 doesn't get a copy of array1 immediately,
as shown by the fact it points to the same address.
Instead, the copy is triggered by the mutation of array2.
This optimization also happens deeper in the structure,
meaning that if for instance your collection is made of other collections,
the latter will also benefit from the copy-on-write mechanism,
as demonstrated by the following snippet (Swift 4.1):
var array1 = [[1, 2], [3, 4]]
var array2 = array1
// Will print the same address twice.
array1[1].withUnsafeBytes { print($0.baseAddress!) }
array2[1].withUnsafeBytes { print($0.baseAddress!) }
array2[0] = []
// Will print the same address as before.
array2[1].withUnsafeBytes { print($0.baseAddress!) }
Replicating copy-on-write
It is in fact rather easy to implement the copy-on-write mechanism in Swift,
as some of the its reference counter API is exposed to the user.
The trick consists of wrapping a reference (e.g. a class instance) within a structure,
and to check whether that reference is uniquely referenced before mutating it.
When that's the case, the wrapped value can be safely mutated,
otherwise it should be copied:
final class Wrapped<T> {
init(value: T) { self.value = value }
var value: T
}
struct CopyOnWrite<T> {
init(value: T) { self.wrapped = Wrapped(value: value) }
var wrapped: Wrapped<T>
var value: T {
get { return wrapped.value }
set {
if isKnownUniquelyReferenced(&wrapped) {
wrapped.value = newValue
} else {
wrapped = Wrapped(value: newValue)
}
}
}
}
var a = CopyOnWrite(value: SomeLargeObject())
// This line doesn't copy anything.
var b = a
However, there is an import caveat here!
Reading the documentation for isKnownUniquelyReferenced we get this warning:
If the instance passed as object is being accessed by multiple threads simultaneously,
this function may still return true.
Therefore, you must only call this function from mutating methods
with appropriate thread synchronization.
This means the implementation presented above isn't thread safe,
as we may encounter situations where it'd wrongly assumes the wrapped object can be safely mutated,
while in fact such mutation would break invariant in another thread.
Yet this doesn't mean Swift's copy-on-write is inherently flawed in multithreaded programs.
The key is to understand what "accessed by multiple threads simultaneously" really means.
In our example, this would happen if the same instance of CopyOnWrite was shared across multiple threads,
for instance as part of a shared global variable.
The wrapped object would then have a thread safe copy-on-write semantics,
but the instance holding it would be subject to data race.
The reason is that Swift must establish unique ownership
to properly evaluate isKnownUniquelyReferenced [4],
which it can't do if the owner of the instance is itself shared across multiple threads.
Value types and multithreading
It is Swift's intention to alleviate the burden of the programmer
when dealing with multithreaded environments, as stated on Apple's blog
(https://developer.apple.com/swift/blog/?id=10):
One of the primary reasons to choose value types over reference types
is the ability to more easily reason about your code.
If you always get a unique, copied instance,
you can trust that no other part of your app is changing the data under the covers.
This is especially helpful in multi-threaded environments
where a different thread could alter your data out from under you.
This can create nasty bugs that are extremely hard to debug.
Ultimately, the copy-on-write mechanism is a resource management optimization that,
like any other optimization technique,
one shouldn't think about when writing code [5].
Instead, one should think in more abstract terms
and consider values to be effectively copied when assigned or passed as arguments.
[1]
This holds only for values used as local variables.
Values used as fields of a reference type (e.g. a class) are also stored in the heap.
[2]
One could get confirmation of that by checking the LLVM byte code that's produced
when dealing with value types rather than reference types,
but the Swift compiler being very eager to perform constant propagation,
building a minimal example is a bit tricky.
[3]
Swift doesn't allow structures to reference themselves,
as the compiler would be unable to compute the size of such type statically.
Therefore, it is not very realistic to think of a structure that is so large
that copying it would become a legitimate concern.
[4]
This is, by the way, the reason why isKnownUniquelyReferenced accepts an inout parameter,
as it's currently Swift's way to establish ownership.
[5]
Although passing copies of value-type instances should be safe,
there's a open issue that suggests some problems with the current implementation
(https://bugs.swift.org/browse/SR-6543).
I don't know if that's the same for every value type in Swift, but for Arrays I'm pretty sure it's a copy-on-write, so it doesn't copy it unless you modify it, and as you said if you pass it around as a constant you don't run that risk anyway.
p.s. In Swift 1.2 there are new APIs you can use to implement copy-on-write on your own value-types too

When do you use obj!.attr and obj?.attr?

If obj not exist obj? generate a nil so obj?.attr too.
If obj is nil then obj!.attr crashes.
But if I am sure obj at that certain point of the code always exist, than for me it seems it is independent which one to use. Am I right? What coding styles do you use?
In my very own opinion, if you're really sure that obj exists, you can use ! or ? either. They produce the same effect on an existing object. The only issue is the compiler: sometimes it's fine to use ! instead of ?, sometimes not.
Anyway, if you want to read further on this, give a chance to the free book by Apple "The Swift Programming Language": these things are very well explained there!
If obj exists, obj?.attr returns an optional type even if attr is not an optional. On the other hand, obj!.attr will be whatever type attr is, so no additional unwrapping is needed if attr is a non-optional.
Consider this example:
class Person {
var age = 37
}
var fred: Person? = Person()
let ageNextYear = fred?.age + 1 // error: value of optional type Int? not unwrapped
let ageLastYear = fred!.age - 1 // this works
You use ? when you create a variable with out giving it a value, meaning it can exist and be used in an unitialized state. And as long as it is not initialized it has no type associated to it.
It has nothing do with the value being nil or not.
As Swift is type safe it requires all variables and constant to always hold a value by default, meaning they have a type. So defining something with ? or ! puts a wrapper around it.
enum Optional {
case None
case Some(T)
}
As you see it either has a type or not. Being nil (if possible for a type) or not has not much to do with it. But everything that does not have a type associated is usually nil.
When you then deal with the value of a variable that was declared as being optional, you need to unwrap it by using ! or else you would use the enumeration showed above.
Unwrapping means that you can assert that it does hold a value meaning it has a type. It takes it out of the enumeration and presents the value it has as a type.
Regarding coding style you usually only need to declare something as ? when you work with C or Objective-C APIs. In pure Swift you will usually not declare something as being optional.
You might need to use ! even in pure swift when something might not be defined. For example you have multi-dimensional array that were initialized as being empty. This is because of how Swift currently handle true multi-dimensional objects where the higher dimensions are implicit optionals.
TL;DR: Don't use ? at all unless you are forced to when dealing with C/Obj-C APIs. When using a value of a variable declared with ? always use ! to refer to the value and never ?.
Some links that explains what happens in more detail:
http://www.drewag.me/posts/what-is-an-optional-in-swift
http://www.drewag.me/posts/uses-for-implicitly-unwrapped-optionals-in-swift