Why I can't use 'object== nil' in Swift? - swift

I have simple method (hope the syntax is right)
func foo() -> WmGroupItemSample?{
var sample:WmGroupItemSample?
if(a > 0){ // some condition
sample = mGroupItemSampleList[0]
}
else{
// do nothing
}
return sample
}
As I understand foo method might return nil if statement a>0 doesn't occur. Therefore return type is: WmGroupItemSample?.
My doubts are in place where I call this method:
var sample:WmGroupItemSample? = foo()
if let temp = sample{
return sample.getStartDate()
}
else{
return -1
}
Is this proper way to validate if not nil? I use Beta 6
What difference if I'll call without ? or with ! like:
var sample:WmGroupItemSample = foo()
or:
var sample:WmGroupItemSample! = foo()
Why I can't validate sample like:
if sample == nil {
/* do one*/
}
else {
}

To compare value with a nil it has to be allowed to be a nil by ? mark. This means that the value can (possibly) be a nil. So by default:
var a : String
can not be a nil. To allow a string be a nil you have to do:
var a : String?
calling variable with ! is unwrapping an optional value
"Trying to use ! to access a non-existent optional value triggers a
runtime error. Always make sure that an optional contains a non-nil
value before using ! to force-unwrap its value." Excerpt From: Apple
Inc. "The Swift Programming Language."
iBooks. https://itun.es/pl/jEUH0.l
For more information I would suggest:
basics
more advanced
reading. Also that SO question with answer might be helpful.

You may have to implement either the Equatable or Comparable protocols.
implementing the Equatable protocol using a global function
func ==<T>(lhs:T, rhs: T?)-> Bool{
return lhs == nil
}
// in your class you can now extend the Equatable protocol in the class definition. ie
class ClassName: Equatable{
// properties and functions
}

Related

Swift where condition to check if a property is implemented

I just found another way to make a great use of protocols and protocol extensions in Swift by extending the Optional protocol to add a function so I can provide default values.
I wrote a blog post about this here: https://janthielemann.de/random-stuff/providing-default-values-optional-string-empty-optional-string-swift-3-1/
The gist of the post is that I needed a clean and easy way to provide default values for optional String which are nil or empty. To do this, I created a Emptyable protocol end extended the Optional protocol like so:
protocol Emptyable {
var isEmpty: Bool { get }
}
extension Optional where Wrapped: Emptyable {
func orWhenNilOrEmpty<T: Emptyable>(_ defaultValue: T) -> T {
switch(self) {
case .none:
return defaultValue
case .some(let value) where value.isEmpty:
return defaultValue
case .some(let value):
return value as! T
}
}
}
extension String: Emptyable {}
Now the question is: Is there a way I can get rid of the Emptyable protocol and instead have a conditional check whether or not a property or function is implemented by the generic type so that I automatically get orWhenNilOrEmpty() for each and every type which has isEmpty?
UPDATE
As suggested by Paulo, the T generic is actually not needed and I created a operator for even quicker access and more convenient usage (at least I think so. Feel free to correct me, I'm always happy to learn new things and improve myself).
I call it the "not empty nil coalescing" operator (who can come up with a better names? I feel like I suck at naming things :/ ). Hopefully some day it helps somebody:
protocol Emptyable {
var isEmpty: Bool { get }
}
infix operator ???: NilCoalescingPrecedence
extension Optional where Wrapped: Emptyable {
func orWhenNilOrEmpty(_ defaultValue: Wrapped) -> Wrapped {
switch(self) {
case .none:
return defaultValue
case .some(let value) where value.isEmpty:
return defaultValue
case .some(let value):
return value
}
}
static func ???(left: Wrapped?, right: Wrapped) -> Wrapped {
return left.orWhenNilOrEmpty(right)
}
}
extension String: Emptyable {}
extension Array: Emptyable {}
extension MyStruct: Emptyable {
let text: String
let number: Int
var isEmpty: Bool { return text.isEmpty && number == 0 }
init(text: String, number: Int) {
self.text = text
self.number = number
}
}
let mandatoryNotEmptyString = optionalOrEmptyString ??? "Default Value"
let mandatoryNotEmptyStruct = optionalOrEmptyStruct ??? MyStruct(name: "Hello World", number: 1)
No, you cannot query if an object or value has a certain property as a constraint on an extension without using a protocol. That would require reflection in a way that is currently not implemented in Swift. Also, an isEmpty property could have different meanings for different types, so testing for the existence of a method or property instead of a protocol could lead to unexpected behaviour.
You could just write
if let unwrappedString = optionalString, !unwrappedString.isEmpty {
// Do stuff
} else {
// Use default value
}
No protocol or extension required and very readable.
In Swift 4, which is coming out this fall, String will conform to BidirectionalCollection, which inherits from Collection. The Collection protocol provides an isEmpty property, so your extension could be
extension Optional where Wrapped: Collection {
// ...
}
But even then you should consider to set empty strings to nil when storing them in the first place, because you now have two states (nil and empty) which seem to represent the exact same thing.

What's wrong with comparing non-optional bool in a single if structure in swift

I have been coding in Swift for a while, and I think I had to put a ! on all my let field variables that were not defined immediately.
Now I notice today that this piece of code does not compile and I am really surprised? why is this?
class MyClass : Mapper {
var a: Bool!
required init?(_ map: Map) {
}
// Mappable
func mapping(map: Map) {
a <- map["a"]
}
}
let myClass = MyClass()
if myClass.a { // Compiler not happy
// Optional type 'Bool!' cannot be used as a boolean; test for '!= nil' instead
}
if true && myClass.a { // Compiler happy
}
if myClass.a && myClass.a { // Compiler happy
}
Apple Swift version 2.2
Edit
Some people point out why am I using a let for a variable that will never change. I mentioned it is for field variables but I shorten the example. When using ObjectMapper (http://github.com/Hearst-DD/ObjectMapper), all the fields are not defined immediately in the init. This is why they are all either optional? or required!
You can declare let a: Bool without a ! and without declaring true or false right then. The compiler will complain if it can't guarantee the value gets set before you use it though.
This works.
let a: Bool
a = true
if a { // Compiler happy
}
This works as well, because a is guaranteed to be set.
let a: Bool
if thingOne < thingTwo {
a = true
} else {
a = false
}
if a { // Compiler happy
}
However, this would not work because a is not guaranteed to be set before you try to use it.
let a: Bool
if thingOne < thingTwo {
a = true
}
if a { // Compiler NOT happy
// "Constant 'a' used before being initialized"
}
Now, if you can't guarantee that your variable will be set by the time you do the check, then you should really be using an optional var in the first place.
var a: Bool?
if a == true { // Compiler happy
}
A bit of history...
In Swift 1.0, it was possible to check if an optional variable optVar contained a value by just checking:
if optVar {
println("optVar has a value")
} else {
println("optVar is nil")
}
In The Swift Programming Language, the update for Swift 1.1 (dated 2014-10-16) stated:
Optionals no longer implicitly evaluate to true when they have a value and false when they do not, to avoid confusion when working with optional Bool values. Instead, make an explicit check against nil with the == or != operators to find out if an optional contains a value.
So, the nonsensical error message that you are getting was put there because the Swift compiler is interpreting your:
if a {
}
to mean:
if a != nil {
}
and it is encouraging you to test against nil to determine if the Optional a has a value.
Perhaps the Swift authors will change it in the future, but for now you will have to explicitly unwrap a:
if a! {
}
or check against true:
if a == true {
}
or (to be completely safe):
if a ?? false {
print("this will not crash if a is nil")
}

Swift - Take Nil as Argument in Generic Function with Optional Argument

I am trying to create a generic function that can take an optional argument.
Here's what I have so far:
func somethingGeneric<T>(input: T?) {
if (input != nil) {
print(input!);
}
}
somethingGeneric("Hello, World!") // Hello, World!
somethingGeneric(nil) // Errors!
It works with a String as shown, but not with nil.
Using it with nil gives the following two errors:
error: cannot invoke 'somethingGeneric' with an argument list of type '(_?)'
note: expected an argument list of type '(T?)'
What am I doing wrong and how should I correctly declare/use this function? Also, I want to keep the usage of the function as simple as possible (I don't want do something like nil as String?).
I guess the compiler can't figure out what T is just from nil.
The following works just fine though for example:
somethingGeneric(Optional<String>.None)
I believe you've overcomplicated the problem by requiring the ability to pass untyped nil (which doesn't really exist; even nil has a type). While the approach in your answer seems to work, it allows for the creation of ?? types due to Optional promotion. You often get lucky and that works, but I've seen it blow up in really frustrating ways and the wrong function is called. The problem is that String can be implicitly promoted to String? and String? can be implicitly promoted to String??. When ?? shows up implicitly, confusion almost always follows.
As MartinR points out, your approach is not very intuitive about which version gets called. UnsafePointer is also NilLiteralConvertible. So it's tricky to reason about which function will be called. "Tricky to reason about" makes it a likely source of confusing bugs.
The only time your problem exists is when you pass a literal nil. As #Valentin notes, if you pass a variable that happens to be nil, there is no issue; you don't need a special case. Why force the caller to pass an untyped nil? Just have the caller pass nothing.
I'm assuming that somethingGeneric does something actually interesting in the case that it is passed nil. If that's not the case; if the code you're showing is indicative of the real function (i.e. everything is wrapping in an if (input != nil) check), then this is a non-issue. Just don't call somethingGeneric(nil); it's a provable no-op. Just delete the line of code. But I'll assume there's some "other work."
func somethingGeneric<T>(input: T?) {
somethingGeneric() // Call the base form
if (input != nil) {
print(input!);
}
}
func somethingGeneric() {
// Things you do either way
}
somethingGeneric(input: "Hello, World!") // Hello, World!
somethingGeneric() // Nothing
Good question and answer. I have an Swift 4 update to contribute:
var str: String? = "Hello, playground"
var list: Array<String>? = ["Hello", "Coder256"]
func somethingGeneric<T>(_ input: T?) {
if (input != nil) {
print(input!);
}
}
func somethingGeneric(_ input: ExpressibleByNilLiteral?) {}
somethingGeneric("Hello, World!") // Hello, World!
somethingGeneric(nil) // *nothing printed*
somethingGeneric(nil as String?) // *nothing printed*
somethingGeneric(str) // Hello, playground
str = nil
somethingGeneric(str) // *nothing printed*
somethingGeneric(list) // ["Hello", "Coder256"]
list = nil
somethingGeneric(list) // *nothing printed*
I figured it out:
func somethingGeneric<T>(input: T?) {
if (input != nil) {
print(input!);
}
}
func somethingGeneric(input: NilLiteralConvertible?) {}
somethingGeneric("Hello, World!") // Hello, World!
somethingGeneric(nil) // *nothing printed*
somethingGeneric(nil as String?) // *nothing printed*
I think that you will never call somethingGeneric(nil) but mostly somethingGeneric(value) or somethingGeneric(function()) for which the compiler has enough info not to be stucked trying to guess the type:
func somethingGeneric<T>(input: T?) {
if let input = input {
print(input);
}
}
func neverString() -> String? {
return nil
}
let a: String? = nil
somethingGeneric("Hello, World!") // Hello, World!
somethingGeneric(a) // Nothing and no error
somethingGeneric(neverString()) // Nothing and no error
Also, I would use the if let syntax instead of if(value != nil).
Here is the solution I came up with that compiles on Swift 5, as many of the solutions here did not compile for me. It might be considered hacky as I use a stored variable to help things along. I was unable to come up with a Swift 5 version of the nil parameters that resolve to type T.
class MyClass {
func somethingGeneric<T>(input: T?) {
if let input = input {
print(input)
}
}
func somethingGeneric() {
somethingGeneric(Object.Nil)
}
}
final class Object {
static var Nil: Object? //this should never be set
}
Actually there is a way to do this, inspired by Alamofire's internal code.
You do not have to install Alamofire to use this solution.
Usage
Your problematic method definition
func someMethod<SomeGenericOptionalCodableType: Codable>(with someParam: SomeGenericOptionalCodableType? = nil) {
// your awesome code goes here
}
What works ✅
// invoke `someMethod` correctly
let someGoodParam1 = Alamofire.Empty.value
someMethod(with: someGoodParam1)
I think it is possible to use Alamofire.Empty.value as a default value in someMethod definition as a parameter.
What does not work ❌
// invoke `someMethod` incorrectly
let someBadParam1: Codable? = nil
let someBadParam2 = nil
someMethod(with: someBadParam1)
someMethod(with: someBadParam2)
Solution definition (source)
/// Type representing an empty value. Use `Empty.value` to get the static instance.
public struct Empty: Codable {
/// Static `Empty` instance used for all `Empty` responses.
public static let value = Empty()
}

Swift Optionals and Forced Unwrapping

I am having a hard time understanding the optionals and forced unwrapping in Swift language. I have read the book and chapters several times but I cannot understand it.
Is there a difference between the following two:
totalAmountTextField?.text.toInt()
totalAmountTextField!.text.toInt()
Also, when declaring the IBOutlets why do I always make it an optional field like this:
#IBOutlet var nameTextField :UITextField?
If I don't use the "?" at the end then it gives errors.
totalAmountTextField?.text.toInt() is equivalent to
func foo() -> Int? { // give you optional Int
if let field = totalAmountTextField {
return field.text.toInt()
} else {
return nil // return nil if totalAmountTextField is nil
}
}
foo()
it should be used if totalAmountTextField can be nil
totalAmountTextField!.text.toInt() is equivalent to
func foo() -> Int { // give you Int
if let field = totalAmountTextField {
return field.text.toInt()
} else {
crash() // crash if totalAmountTextField is nil
}
}
foo()
it should be used only if you know totalAmountTextField must not be nil
// It represents that totalAmountTextField may be nil and then stop the chain.
totalAmountTextField?.text.toInt()
// It assume that totalAmountTextField must have value, if not then caused a crash.
totalAmountTextField!.text.toInt()
You can take a look at the Swift Documentation about Optional Chaining. https://developer.apple.com/library/prerelease/mac/documentation/Swift/Conceptual/Swift_Programming_Language/OptionalChaining.html

Protocol: Cannot assign to 'X' in 'Y' in Swift

I just defined a very simple protocol and a a class using generics which can handle this protocol.
In the lines marked with error you will get the error: "Cannot assign to 'flag' in 'aObj'.
protocol Flag {
var flag: Bool {get set}
}
class TestFlag<T: Flag> {
func toggle(aObj: T) {
if aObj.flag {
aObj.flag = false; // <--- error
} else {
aObj.flag = true; // <--- error
}
}
}
Do you have an idea why and what I have to change to fix it?
From the docs:
Function parameters are constants by default. Trying to change the
value of a function parameter from within the body of that function
results in a compile-time error. This means that you can’t change the
value of a parameter by mistake.
In this case, you can add inout so that the toggle is persisted beyond your function call:
func toggle(inout aObj: T) {
if aObj.flag {
aObj.flag = false;
else {
aObj.flag = true;
}
}
You could have also done:
func toggle(var aObj: T) {
}
but that might not achieve what you wanted.
manojlds answer is correct and therefore I accepted it.
Nevertheless there was a similar answer some days ago with the same solution but with a other argumentation (seems now deleted).
The argumentation was about that the compliler can not know if the protocol is used for a class, a struct or a enum. With Swift, protocols can by applied on all this types. But struct instances use a by-value call and for classes instances (objects) it us a by-reference call.
From my perspective this answer was correct too, because you can solve the problem with a 2nd solution:
#objc
protocol Flag {
var flag: Bool {get set}
}
Just add the #obj attriute on the protocol. As a result you can use this protocol only for a class which lead to the result only by-refernece calls are allowd. Therefore the compiler don't need anymore the inout information.
But I searched for a solution to increase the reuse of the protocol and use now manojlds suggestions.