I am looking for the way to write short syntax.
For instance. In JS, PHP and etc.
var a = 1 ;
function Foo ()-> void {}
a && Foo() ;
if a exists, run Foo.
a and Foo itself already mean exist or not, the syntax is away better looks....
However, in Swift, the typing checking is kinda of tough.
var a = 1 ;
func Foo ()-> Foid {} ;
a && Foo();
will generate neither are Bool returning error.
a != nil && Foo() ;
this can resolve and variable condition, but what if the better bypass for the function condition? I just dont want to write something like
if( a != nil ) { Foo() } ;
Yet what is the better syntax for Not Exist?
if ( !a ) or !a //is easy and better looks...
I found not similar thing in swift...
if( a == nil ) // will throws error when its not Optional Typing.
guard var b = xxx else {} // simply for Exist and very long syntax.
Thank you for your advice!
As mentioned by other contributors, Swift emphasizes readability and thus, explicit syntax. It would be sacrilege for the Swift standard library to support Python-style truth value testing.
That being said, Swift’s extensibility allows us to implement such functionality ourselves—if we really want to.
prefix func !<T>(value: T) -> Bool {
switch T.self {
case is Bool.Type:
return value as! Bool
default:
guard Double(String(describing: value)) != 0
else { return false }
return true
}
}
prefix func !<T>(value: T?) -> Bool {
guard let unwrappedValue = value
else { return false }
return !unwrappedValue
}
var a = 1
func foo() -> Void { }
!a && !foo()
Or even define our own custom operator:
prefix operator ✋
prefix func ✋<T>(value: T) -> Bool {
/// Same body as the previous example.
}
prefix func ✋<T>(value: T?) -> Bool {
guard let unwrappedValue = value
else { return false }
return ✋unwrappedValue
}
var a = 1
func foo() -> Void { }
✋a && ✋foo()
The expectations you've developed from dynamic languages like PHP and JS (and Ruby, Python for that matter) are almost universally inapplicable to static languages like Swift.
Swift is a statically compiled language. If you reference a variable that doesn't exist, it's not legal Swift code, and the compiler will fail your build. Given that, the question of "how do I check if a variable is undefined?" is completely moot in Swift. If you have a successfully compiling program that references a variable a, then a exists. There's absolutely no reason for a check, and so a mechanism for it doesn't even exist.
Static vs Dynamic typing
Static type systems are like mathematical proof systems. They produce rigerous proofs that certain aspects of your program are valid. This has trade-offs. The rigidity buys you many guarantees. For example, you'll never have a valid Swift program where you accidentally pass an Int where a Bool is expected. The static type system makes that class of error literally impossible, so it's not something you have to remember to check for yourself.
On the other hand, many truths are easier to intuit than to prove. Thus, there's great utility in scripting and dynamic languages, because they don't demand the rigorous proofs of your claims that static languages require. On the down side, their type systems "do" much less. For example, JS happily lets you reference an undefined variable. To remedy this, JS provides a way for you to do a run-time check to see whether a variable is defined or not. But this isn't a problem Swift has, so the "solution" is absent.
When static typing is too hard
Swift actually takes a middle ground position. If you find yourself with a statement that's obviously true, but hard to prove to the compiler, various "escape hatches" exist that allow you to leave the safety of the type system, and go into dynamic land. For example, if you look at an IBOutlet, and see that it's connected to an element in a storyboard, you can intuitively be sure that the IBOutlet is not nil. But that's not something you can prove to the compiler, and hence when you see implicitly unwrapped optionals being used for IBOutlets.
Implicitly unwrapped optionals are one such "escape hatch". The Any type is another, as is unsafeBitcast(_:to:), withoutActuallyEscaping(_:), as!, try!, etc.
Swift takes type safety very seriously. Unlike C or JS we can not use anything that doesn't resolve to Bool value type in If statement in Swift. So there won't be a short hand for that(at-least that I know of). Regarding below code
if( a == nil ) // will throws error when its not Optional Typing.
Swift doesn't allow you to set nil to non optional types. So there is no need to check for nil. By the way both Obj-C and Swift use verbose syntax, we need to get use to that.
In this case you are trying to force Swift to work in a way that you are used to with other languages like JavaScript or PHP, as you say in your comment. There are a few reasons why your code won't compile, but it mainly falls on the fact that Swift doesn't do the same truthy and falsy stuff JS does.
var a = 1
if a {
print("won't compile")
}
//'Int' is not convertible to 'Bool'
In Swift it's better to use an actual Bool value if that's what it's supposed to be, or if it's truly supposed to be an Int you're just going to have to check the value
var a = true
if a {
print("this compiles")
}
or
var a = 1
if a > 0 {
print("this compiles too")
}
Swift really isn't meant to be as loose as JS, so you should just embrace that and take advantage of the safety and readability.
Here is one way most similar to what you designed.
You may have to set the type of a to Int?:
var a: Int? = 1
func foo ()-> Void {}
a.map{_ in foo()}
Related
This article shows: ?? is very time-consuming, and the test found it to be true. So I want to optimize this:
#if DEBUG
public func ?? <T>(left: T?, right: T) -> T {
guard let value = left else {
return right
}
return value
}
#endif
BUT
string = string ?? ""
ERROR: Ambiguous use of operator '??'
The article you link to does not describe how the nil coalescing operator is "time consuming", but states the simple fact that a compound expression of two nil coalescing operator calls as well as other evaluations has a significantly longer build time than if we break down this compound expressions into smaller parts, especially if the former contains lazy evaluations at some point (which is the case e.g. for the evaluation of the rhs of the ?? operator). By breaking down complex expression we help the compiler out; Swift has been known to have some difficulty compiling complex compound statements, so this is expected. This, however, should generally not affect runtime performance, given that you build your implementation for release mode.
By overloading the implementation of the ?? operator during debug with a non-lazy evaluation of the lhs while still using long compound statements shouldn't fully redeem the issue described in the previous clause. If compile time is really an issue for you, use the same approach as the author of the article; break down your complex expressions into several smaller ones, in so doing some of the work of the compiler yourself. Refactoring with the sole purpose of decreasing compile times (which we might believe is the job of the compiler, not us) might be frustrating at times, but is an issue we (currently) has to live with when working with Swift. See e.g. the following excellent article covering the Swift compiler issues with long/complex expressions:
Exponential time complexity in the Swift type checker
W.r.t. to the ambiguity error: for reference, the official implementation of the nil coelescing operator can be found here:
#_transparent
public func ?? <T>(optional: T?, defaultValue: #autoclosure () throws -> T)
rethrows -> T {
switch optional {
case .some(let value):
return value
case .none:
return try defaultValue()
}
}
And we may note that it has the same specificity as your own generic implementation (compare the signatures!), meaning the ambiguity error is to be expected. E.g., compare with:
// due to the '#autoclosure', these two have the
// same precedecence/specificity in the overload
// resolution of 'foo(_:)'
func foo<T>(_ obj: T) {}
func foo<T>(_ closure: #autoclosure () -> T) {}
foo("foo") // ERROR at line 9, col 1: ambiguous use of 'foo'
When you implement your ?? overload it as in your answer, typing out Any, this implementation becomes more specific than the generic one, which means it will take precedence in the overload resolution of ??. Using Any is such contexts, however, is generally a bad idea, attempting to mimic dynamic typing rather than relying on Swift's renowned static typing.
if I overloading ??,I can't use <T>...
then
#if DEBUG
public func ?? (left: Any?, right: Any) -> Any {
guard let value = left else {
return right
}
return value
}
#endif
is OK!
BUT I can't get types of return value... 😭
I will first explain what I'm trying to do and how I got to where I got stuck before getting to the question.
As a learning exercise for myself, I took some problems that I had already solved in Objective-C to see how I can solve them differently with Swift. The specific case that I got stuck on is a small piece that captures a value before and after it changes and interpolates between the two to create keyframes for an animation.
For this I had an object Capture with properties for the object, the key path and two id properties for the values before and after. Later, when interpolating the captured values I made sure that they could be interpolated by wrapping each of them in a Value class that used a class cluster to return an appropriate class depending on the type of value it wrapped, or nil for types that wasn't supported.
This works, and I am able to make it work in Swift as well following the same pattern, but it doesn't feel Swift like.
What worked
Instead of wrapping the captured values as a way of enabling interpolation, I created a Mixable protocol that the types could conform to and used a protocol extension for when the type supported the necessary basic arithmetic:
protocol SimpleArithmeticType {
func +(lhs: Self, right: Self) -> Self
func *(lhs: Self, amount: Double) -> Self
}
protocol Mixable {
func mix(with other: Self, by amount: Double) -> Self
}
extension Mixable where Self: SimpleArithmeticType {
func mix(with other: Self, by amount: Double) -> Self {
return self * (1.0 - amount) + other * amount
}
}
This part worked really well and enforced homogeneous mixing (that a type could only be mixed with its own type), which wasn't enforced in the Objective-C implementation.
Where I got stuck
The next logical step, and this is where I got stuck, seemed to be to make each Capture instance (now a struct) hold two variables of the same mixable type instead of two AnyObject. I also changed the initializer argument from being an object and a key path to being a closure that returns an object ()->T
struct Capture<T: Mixable> {
typealias Evaluation = () -> T
let eval: Evaluation
let before: T
var after: T {
return eval()
}
init(eval: Evaluation) {
self.eval = eval
self.before = eval()
}
}
This works when the type can be inferred, for example:
let captureInt = Capture {
return 3.0
}
// > Capture<Double>
but not with key value coding, which return AnyObject:\
let captureAnyObject = Capture {
return myObject.valueForKeyPath("opacity")!
}
error: cannot invoke initializer for type 'Capture' with an argument list of type '(() -> _)'
AnyObject does not conform to the Mixable protocol, so I can understand why this doesn't work. But I can check what type the object really is, and since I'm only covering a handful of mixable types, I though I could cover all the cases and return the correct type of Capture. Too see if this could even work I made an even simpler example
A simpler example
struct Foo<T> {
let x: T
init(eval: ()->T) {
x = eval()
}
}
which works when type inference is guaranteed:
let fooInt = Foo {
return 3
}
// > Foo<Int>
let fooDouble = Foo {
return 3.0
}
// > Foo<Double>
But not when the closure can return different types
let condition = true
let foo = Foo {
if condition {
return 3
} else {
return 3.0
}
}
error: cannot invoke initializer for type 'Foo' with an argument list of type '(() -> _)'
I'm not even able to define such a closure on its own.
let condition = true // as simple as it could be
let evaluation = {
if condition {
return 3
} else {
return 3.0
}
}
error: unable to infer closure type in the current context
My Question
Is this something that can be done at all? Can a condition be used to determine the type of a generic? Or is there another way to hold two variables of the same type, where the type was decided based on a condition?
Edit
What I really want is to:
capture the values before and after a change and save the pair (old + new) for later (a heterogeneous collection of homogeneous pairs).
go through all the collected values and get rid of the ones that can't be interpolated (unless this step could be integrated with the collection step)
interpolate each homogeneous pair individually (mixing old + new).
But it seems like this direction is a dead end when it comes to solving that problem. I'll have to take a couple of steps back and try a different approach (and probably ask a different question if I get stuck again).
As discussed on Twitter, the type must be known at compile time. Nevertheless, for the simple example at the end of the question you could just explicitly type
let evaluation: Foo<Double> = { ... }
and it would work.
So in the case of Capture and valueForKeyPath: IMHO you should cast (either safely or with a forced cast) the value to the Mixable type you expect the value to be and it should work fine. Afterall, I'm not sure valueForKeyPath: is supposed to return different types depending on a condition.
What is the exact case where you would like to return 2 totally different types (that can't be implicitly casted as in the simple case of Int and Double above) in the same evaluation closure?
in my full example I also have cases for CGPoint, CGSize, CGRect, CATransform3D
The limitations are just as you have stated, because of Swift's strict typing. All types must be definitely known at compile time, and each thing can be of only one type - even a generic (it is resolved by the way it is called at compile time). Thus, the only thing you can do is turn your type into into an umbrella type that is much more like Objective-C itself:
let condition = true
let evaluation = {
() -> NSObject in // *
if condition {
return 3
} else {
return NSValue(CGPoint:CGPointMake(0,1))
}
}
In c++, one can introduce an alias reference as follows:
StructType & alias = lengthyExpresionThatEvaluatesToStuctType;
alias.anAttribute = value; // modify "anAttribute" on the original struct
Is there a similar syntactic sugar for manipulating a (value typed) struct in Swift?
Update 1: For example: Let say the struct is contained in a dictionary of kind [String:StructType], and that I like to modify several attributes in the the struct myDict["hello"]. I could make a temporary copy of that entry. Modify the copy, and then copy the temporary struct back to the dictionary, as follows:
var temp = myDict["hello"]!
temp.anAttribute = 1
temp.anotherAttribute = "hej"
myDict["hello"] = temp
However, if my function has several exit points I would have to write myDict["hello"] = temp before each exit point, and it would therefore be more convinient if I could just introduce and alias (reference) for myDict["hello"] , as follows:
var & alias = myDict["hello"]! // how to do this in swift ???
alias.anAttribute = 1
alias.anotherAttribute = "hej"
Update 2: Before down- or close- voting this question: Please look at Building Better Apps with Value Types in swift (from WWWDC15)!! Value type is an important feature of Swift! As you may know, Swift has borrowed several features from C++, and value types are maybe the most important feature of C++ (when C++ is compared to Java and such languages). When it comes to value types, C++ has some syntactic sugar, and my questions is: Does Swift have a similar sugar hidden in its language?. I am sure Swift will have, eventually... Please, do not close-vote this question if you do not understand it!
I have just read Deitel's book on Swift. While I'am not an expert (yet) I am not completely novel. I am trying to use Swift as efficient as possible!
Swift doesn't allow reference semantics to value types generally speaking, except when used as function parameters declared inout. You can pass a reference to the struct to a function that works on an inout version (I believe, citation needed, that this is implemented as a copy-write, not as a memory reference). You can also capture variables in nested functions for similar semantics. In both cases you can return early from the mutating function, while still guaranteeing appropriate assignment. Here is a sample playground that I ran in Xcode 6.3.2 and Xcode 7-beta1:
//: Playground - noun: a place where people can play
import Foundation
var str = "Hello, playground"
struct Foo {
var value: Int
}
var d = ["nine": Foo(value: 9), "ten": Foo(value: 10)]
func doStuff(key: String) {
let myNewValue = Int(arc4random())
func doMutation(inout temp: Foo) {
temp.value = myNewValue
}
if d[key] != nil {
doMutation(&d[key]!)
}
}
doStuff("nine")
d // d["nine"] has changed... unless you're really lucky
// alternate approach without using inout
func doStuff2(key: String) {
if var temp = d[key] {
func updateValues() {
temp.value = Int(arc4random())
}
updateValues()
d[key] = temp
}
}
doStuff2("ten")
d // d["ten"] has changed
You don't have to make the doMutation function nested in your outer function, I just did that to demonstrate the you can capture values like myNewValue from the surrounding function, which might make implementation easier. updateValues, however, must be nested because it captures temp.
Despite the fact that this works, based on your sample code, I think that using a class here (possibly a final class if you are concerned about performance) is really more idiomatic imperative-flavored Swift.
You can, if you really want to, get a raw pointer using the standard library function withUnsafeMutablePointer. You can probably also chuck the value into an inner class that only has a single member. There are also functional-flavored approaches that might mitigate the early-return issue.
accodring to apple documentation the following code checks if convertedNumber has been initialized, and if so, it executes the if block.
let convertedNumber: Int?
if convertedNumber {
//do stuff
}
I'm interested in what's happening behind the scenes. Is this some sort of shorthand notation, because the condition in if statement must evaluate to boolean true or false. how is the fact that convertedNumebr contains a value transformed into true or false?
I was thinking that this is a shorthand notation for:
if convertedNumber!=nil {
//do stuff
}
correct me if I'm wrong.
The "shorthand test" for optionals
if someOptional {
//do stuff
}
existed only in early version of Swift.
From the Xcode 6 beta 5 release notes:
Optionals no longer conform to the BooleanType (formerly LogicValue)
protocol, so they may no longer be used in place of boolean
expressions (they must be explicitly compared with v != nil). This
resolves confusion around Bool? and related types, makes code more
explicit about what test is expected, and is more consistent with the
rest of the language.
So with current Xcode versions, you have to test with
if someOptional != nil {
//do stuff
}
or, of course, with optional binding.
In addition, struct Optional conforms to the NilLiteralConvertible
protocol, so that (in this context) nil is identical to
Optional<T>.None with the matching type T.
I'm writing a program in Swift which has various functions which should always return values, but in some cases cannot, and so the program crashes and alerts the user of the error. To accomplish this, I am using implicitly unwrapped Optionals as return values for my functions, so that error cases can just return nil (even though that code will never be reached). It looks something like this:
func giveMeAThing() -> Thing! {
if !possibleErrorCase() {
var thing = createAThingForSure()
return thing
}
else {
crashAndBurn()
return nil
}
}
Writing this, it feels a bit hacky, so my question is whether or not this is a good use of implicitly unwrapped Optionals in Swift. Is there a better way to structure my code that will avoid this?
It may be that the piece of the puzzle you are missing is #noreturn. If you declare a function as #noreturn, the compiler abandons any complaints if you don't fulfill the surrounding contract. Thus you can call such a method and not return any value, and the compiler will not complain because it knows we are going to die anyway. For example, fatalError and abort are declared this way.
Thus, the following is legal:
func f() -> String {
fatalError("oooooops")
}
That compiles even though we fail to fulfill the contract of returning a value, because #noreturn (in the declaration of fatalError) tears up the contract and flushes the pieces down the toilet.
Another possible way of approaching the same issue is to use assert. This crashes and burns for you if a condition is not met. So, in your example:
func giveMeAThing() -> Thing {
assert(!possibleErrorCase(), "gaaaaah!") // crash and burn if error case
return createAThingForSure()
}
The downside is that by default all asserts succeed in a shipping app. But it's great during development, and after all your shipping app should not be crashing and burning anyway.
I'd rather write this as
func giveMeAThing() -> Thing! {
if possibleErrorCase() {
crashAndBurn() // I assume this will not return anyhow
}
return createAThingForSure()
}
but that's likely just a question of taste.
You don't need an implicitly unwrapped optional since you don't expect the second case to return any value.
As commenters noted, use fatalError in the fail case.