I am quite new to Swift and when I learning about initializing a string, I find a wired syntax that I cannot understand.
For example
If I initialize a string using:
var str:String = "Hello, playground"
str.isEmpty
This works well
However, if I initialize a string with a constructor
var str = String("Hello, playground")
str.isEmpty
this does not work.
And the compiler fix it by changing the syntax to
str?.isEmpty
I have no idea about what is that “?” for.
Any suggestion is appreciated.
When you say:
let str = String("Hello, playground")
you're using String's init?(_ description: String) initialiser, which satisfies LosslessStringConvertible's initialiser requirement, as not all types that conform have a representation for an arbitrary string (for example, Double).
Because the initialiser is failable, it returns an optional string instance (i.e String?), hence why the compiler is prompting you to use optional chaining (if not already, I would highly recommend reading the optionals section of the language guide).
However it's worth noting there's absolutely no need for String's implementation of this initialiser to be failable – as a string can always represent a string! This will be fixed in Swift 4, it will be implemented as a non-failable initialiser satisfying the failable requirement. This is because it doesn't break the contract with the protocol of either returning a new instance or nil (it just never happens to do the latter).
In the mean time however, you could just force unwrap the result, as it will always succeed. But really the use of the initialiser here is completely redundant. You should use a string literal, such as in your first example:
let str = "Hello, playground"
str will simply be initialised to a non-optional String instance with the contents of the string literal. Note that Swift can infer str to be of type String – you don't need to explicitly annotate it as such.
var str = String("Hello, playground")
This produces what is called an "Optional". In Swift, if something can be null/nil, then it is wrapped in an Optional. The '?' tries to unwrap the optional and produce a String.
If you KNOW that the value can never be null, then you can force unwrap it like this:
var str = String("Hello, playground")
str!.isEmpty
or
var str2 = String("Hello, playground")!
str2.isEmpty
Generally though forced unwrapping is frowned upon, as it can lead to crashes. Better approaches:
struct MyError:Error {
var message:String
}
guard let str3 = String("Hello, playground") else {
throw MyError(message:"failed")
}
str3.isEmpty
if let str4 = String("Hello, playground") {
str4.isEmpty
}
Related
var tuples: String! = "subash"
if let goin = tuples {
print(goin!)
}
I am receiving this error:
Cannot force unwrap the value of non-optional type String
I don't what happening constant goin is same as tuples but why it's showing me an error when I do force unwrap
Instead of the above code, this is running well:
var tuples: String! = "subash"
print(tuples!)
But kindly I need a solution to my above problem
That's something normal if you know how optionals work.
Inside the if statement, the right expression must be either Optional or Implicitly Unwrapped Optional, BUT NOT "normal" value.
This is the correct code:
let tuples: String! = "subash"
if let goin = tuples {
print(goin) // NO NEED to unwrap, because going IS NOT Optional
}
The reason this code runs fine:
var tuples: String! = "subash"
print(tuples!)
print(tuples)
... is because tuples is of type Implicitly Unwrapped Optional.
However, in general cases like this:
let myVar: String! = "some string"
if let myNewVar = myVar {
// Some code...
}
... myVar is always an Implicitly Unwrapped Optional, whereas myNewVar is of String type, because of how Optional Unwrapping works with if let statements.
Finally, if we unwrap the value this way:
let myVar: String! = "some string"
if let myVar = myVar {
// Some code...
print(myVar)
}
The printed value is the temp myVar, which is of type String and shadows the Implicitly Unwrapped myVar variable we initially declare.
Why are implicitly unwrapped optionals not unwrapped when using string interpolation in Swift 3?
Example:
Running the following code in the playground
var str: String!
str = "Hello"
print("The following should not be printed as an optional: \(str)")
produces this output:
The following should not be printed as an optional: Optional("Hello")
Of course I can concatenate strings with the + operator but I'm using string interpolation pretty much everywhere in my app which now doesn't work anymore due to this (bug?).
Is this even a bug or did they intentionally change this behaviour with Swift 3?
As per SE-0054, ImplicitlyUnwrappedOptional<T> is no longer a distinct type; there is only Optional<T> now.
Declarations are still allowed to be annotated as implicitly unwrapped optionals T!, but doing so just adds a hidden attribute to inform the compiler that their value may be force unwrapped in contexts that demand their unwrapped type T; their actual type is now T?.
So you can think of this declaration:
var str: String!
as actually looking like this:
#_implicitlyUnwrapped // this attribute name is fictitious
var str: String?
Only the compiler sees this #_implicitlyUnwrapped attribute, but what it allows for is the implicit unwrapping of str's value in contexts that demand a String (its unwrapped type):
// `str` cannot be type-checked as a strong optional, so the compiler will
// implicitly force unwrap it (causing a crash in this case)
let x: String = str
// We're accessing a member on the unwrapped type of `str`, so it'll also be
// implicitly force unwrapped here
print(str.count)
But in all other cases where str can be type-checked as a strong optional, it will be:
// `x` is inferred to be a `String?` (because we really are assigning a `String?`)
let x = str
let y: Any = str // `str` is implicitly coerced from `String?` to `Any`
print(str) // Same as the previous example, as `print` takes an `Any` parameter.
And the compiler will always prefer treating it as such over force unwrapping.
As the proposal says (emphasis mine):
If the expression can be explicitly type checked with a strong optional type, it will be. However, the type checker will fall back to forcing the optional if necessary. The effect of this behavior is that the result of any expression that refers to a value declared as T! will either have type T or type T?.
When it comes to string interpolation, under the hood the compiler uses this initialiser from the _ExpressibleByStringInterpolation protocol in order to evaluate a string interpolation segment:
/// Creates an instance containing the appropriate representation for the
/// given value.
///
/// Do not call this initializer directly. It is used by the compiler for
/// each string interpolation segment when you use string interpolation. For
/// example:
///
/// let s = "\(5) x \(2) = \(5 * 2)"
/// print(s)
/// // Prints "5 x 2 = 10"
///
/// This initializer is called five times when processing the string literal
/// in the example above; once each for the following: the integer `5`, the
/// string `" x "`, the integer `2`, the string `" = "`, and the result of
/// the expression `5 * 2`.
///
/// - Parameter expr: The expression to represent.
init<T>(stringInterpolationSegment expr: T)
Therefore when implicitly called by your code:
var str: String!
str = "Hello"
print("The following should not be printed as an optional: \(str)")
As str's actual type is String?, by default that's what the compiler will infer the generic placeholder T to be. Therefore the value of str won't be force unwrapped, and you'll end up seeing the description for an optional.
If you wish for an IUO to be force unwrapped when used in string interpolation, you can simply use the force unwrap operator !:
var str: String!
str = "Hello"
print("The following should not be printed as an optional: \(str!)")
or you can coerce to its non-optional type (in this case String) in order to force the compiler to implicitly force unwrap it for you:
print("The following should not be printed as an optional: \(str as String)")
both of which, of course, will crash if str is nil.
I have the following variable:
var npill : String!
It's an Int value, but I can't set it as Int because of:
npillIn: fieldNumeroPillole.text!,
How can I convert this var to a Int var? I have tried the following:
var number1: Int = (npill! as NSString).intValue
By the above code I receive the following error:
cannot use instance member 'npill' within property initializer, property initializers run before "self" is aviable
If I then set:
var number1: Int = (self.npill! as NSString).intValue
The error it outputs is as follows:
Value of type '(NSObject) -> () -> Farmaco' has no member 'npill'
If anyone knows how I should be converting it properly, please let me know.
Update
Thank you to #Hamish for pointing out what the OP was asking
So the problem seems to be this
import Foundation
class Foo {
var npill : String!
var number1: Int = (npill! as NSString).intValue
}
error: cannot use instance member 'npill' within property initializer; property initializers run before 'self' is available
var number1: Int = (npill! as NSString).intValue
^
What's going on here?
You are using a property to populate another property, and this is not allowed.
Solution
However you can easily fix the problem postponing the initialisation of number1. Infact if you make number1 lazy, it will be populated only when used.
class Foo {
var npill : String!
lazy var number1: Int = { return Int(self.npill!)! }()
}
Warning: Of course this code will crash if npill is still nil when number1 is used.
Old version
You can simply write
let npill: String! = "34"
if let npill = npill, let num = Int(npill) {
print(num) // <-- here you have your Int
}
(As #Hamish pointed out in a comment below, I misunderstood what the OP was really asking about. I'll leave my answer, however, as some curiosa and insights regarding ! type annotation, which may be relevant for future readers of this question)
For any type of String optionals, their values needs to be unwrapped prior to using the failable init?(_ text: String) initializer or Int.
In your example, the variable npill is an optional, as you've annotated its type with the ! specifier (which should be used with care). Quoting from the implemented evolution proposal SE-0054 [emphasis mine]
Appending ! to the type of a Swift declaration will give it optional
type and annotate the declaration with an attribute stating that it
may be implicitly unwrapped when used.
Hence, it's entirely legal to use npill directly with the init?(_ text: String) initializer of Int, as it will be unwrapped (without any safety check for nil content!) on-the-fly upon use.
// UNSAFE example!
var npill: String! = "42"
if let npillInt = Int(npill) {
/* ^^^^^^^^ ^^^^^- since 'npill' has a type annotated with
| '!', it will be unsafely unwrapped at
| this point
\
the optional binding here safely unwraps the return from
the failable Int initializer, but has nothing to do with
the unwrapping of 'npill' */
print(npillInt) // 42
}
// why unsafe? consider
npill = nil
if let npillInt = Int(npill) { // runtime exception!
// ...
}
Generally you should avoid using the ! annotation, however, unless you are entirely certain that the content of the resulting optional variable will never ever be nil.
Leaving aside the cons of even using the ! annotation: you may implement a safe version of the unsafe example above, by overriding the unsafe implicit unwrapping with safe explicit unwrapping techniques. For a given optional variable declared using the ! annotation, we may still apply safe means to unwrap it, e.g. optional binding or using the nil coalescing operator. #appzYourLife has already showed one perfectly valid and safe way to handle the unwrapping and attempted type conversion of npill using optional binding, so I'll simply include another example using the nil coalescing operator instead:
// "safe" example (STILL: why use the `!` annotation?)
var npill: String! = "42"
if let npillInt = Int(npill ?? "x") {
/* ^^^^^ if 'npill' is 'nil', the Int initializer will
be given the value "x", which itself will lead
it to fail, which is safe here as we intend to
bind the result of the initialization to 'npillInt' */
print(npillInt) // 42
}
npill = nil
if let npillInt = Int(npill ?? "x") {
// ... doesnt enter
}
The consensus of the examples above is that if we're even slightly uncertain whether npill can ever be nil or not, we need to treat it as if it was just an optional not type annotated with ! (i.e. String?); overriding the default unsafe unwrapping with safe means when working with the variable. In such a case, why would we even want to use the ! typ annotation at all, as it only brings fragility/danger to our application?
I understand the difference between String! type and String? type. But what about String type? How does it differs from String! and String? in swift? Does String! type identical to String type?
Say, I have a class like this:
class Person {
private var _name: String!
var name: String {
return _name
}
init(name: String) {
_name = name
}
}
There is no compiler error, looks like String type is identical to String! type. But I am not sure...
String and String! are not identical. There is just happens to be enough sugar in the language to convert between them. Similarly there is sugar in the language to convert between String and String? (but not in reverse).
Start with the basics. There is String. Unless there is some strong reason, you should use String when you mean a string of characters. Everything else is "more stuff" and you shouldn't add it unless you need it.
There is Optional<String>. This is just an enum with two cases, one with a value, and one without a value:
public enum Optional<Wrapped> : _Reflectable, NilLiteralConvertible {
case None
case Some(Wrapped)
// ...
}
There is a postfix operator ! for Optional which will return Wrapped if it's available, and crash if it is not. So far, no magic. This is stuff you could build yourself.
There are are few pieces of magic around Optional. First, the type Wrapped? is magically converted to Optional<Wrapped> by the compiler (for any Wrapped). This is just syntactic sugar. The two notations are identical. Second, there is optional-chaining with the ?. "operator" (it's not really an operator; it's part of the language and you couldn't build it yourself). And then there's optional promotion. Any type Wrapped can be automatically and implicitly converted to Wrapped? if needed. There are are few other pieces of magic around Optional like if-let syntax, and there's nil which is a synonym for Optional.None (I believe they're actually identical). But Optional really is just a generic type, implemented as an enum. It's just a type the compiler knows special things about.
Then there is ImplicitlyUnwrappedOptional<Wrapped>. This is also just an enum (in Swift 2.2; this will change in Swift 3).
public enum ImplicitlyUnwrappedOptional<Wrapped> : _Reflectable, NilLiteralConvertible {
case None
case Some(Wrapped)
// ...
}
This is not the same as Optional and it is not the same as Wrapped. It's a completely different type. But it also has some magic associated with it. First, the type Wrapped! is syntactic sugar for ImplicitlyUnwrappedOptional<Wrapped>. Again, it's just sugar. The two are the same (in Swift 2.2, not in Swift 3). Next, if IUO<Wrapped> is found in a place that Wrapped is expected, it will automatically be converted to Wrapped or crash if there is no value. If it is found in a place that Wrapped? is expected, it will automatically be converted to Wrapped?. Those are magical, and it's why sometimes String and String! seem to be the same type. That's just the compiler magically "making it work" for you by adding an invisible conversion step. It doesn't mean they're really the same type.
IUO is mostly useful in bridging to certain Objective-C patterns, especially involving Storyboards, and should be avoided outside of those situations. Even in those situations, IUO is just there for convenience. You could do all the same things using regular Optionals, you'd just have to check them with if-let more often. Using Optionals is much safer than using IUO. It's easy to think "I know for certain this value will always be set before it is used." And just this week I chased down a crasher due to being wrong about that. There's a difference between "it should be" and "it must be." But, being totally safe with Optionals in Storyboards could be very inconvenient and might mask some bugs (by doing nothing rather than crashing), so that's the most common place for IUO.
IUO properties used to be valuable for dealing with failable init methods. That's no longer a problem in Swift 2.2, so that use has gone away. The last "pure Swift" use that I run into is when you must pass self to the initializer of something you store as a property (you can't pass self at that point because all your properties have been initialized). That's an unfortunate use-case and very messy, and I hope we come up with a fix for it. Outside of these kinds of cases, you should avoid Implicitly Unwrapped Optionals.
String and String! are same and different at the same time. If you declare let or var as String, you will have to set some value at init method. If you declare it as String! you can set value when you want but you must do it before reading this value.
So, if your _name will be nil, your app will crash.
If you declare it as String, compiler guarantees that _name != nil before the first reading. If you declare it as String!, you should guarantee it.
This things have same type because of this:
var test1: String = "test"
var test2: String! = "test"
if test1 is String { // True
print("test1 is String")
}
if test2 is String { // True
print("test2 is String")
}
This types are equal but this things are different
Note the difference between these two functions:
func f() {
let x: String? = nil
guard let y: String = x else {
print("failed")
return
}
}
func g() {
let x: String? = nil
guard let y: String! = x else {
print("failed")
return
}
print(y.dynamicType, y)
}
f() # failed
g() # Optional<String> nil
You are better to read swift documentation than any answer here. Properly undestandig swift's type system is very important, so do it first.
both String! and String? are optional types.
Using String? requires you to test whether the value is nil (.None) or a value (.Some). For example using guard statements or if-let clauses.
String! is an optional that is assumed to be non-nil and is unwrapped automatically. This allows you to write cleaner-looking code and avoid unwrapping wherever the value is used. BUT: if your value is nil a fatal error will occur.
I have a string var oneString: String! and later on in a method, when I want to concatenate a string to oneString I have to do this:
oneString! += anyString
If I don't add ! I get an error 'String!' is not identical to 'CGFloat'
If I initialize my string with var oneString = "" I don't have this problem. Why? Why do I need to unwrap oneString while I explicitly said it would not be nil when I declared it?
Why do I need to unwrap oneString while I explicitly said it would not be nil when I declared it?
You’ve misunderstood what var oneString: String! means. It does not mean oneString will not be nil. If you declare a type as var oneString: String, then you are declaring a type that cannot be nil.
The type String! is an “implicitly-unwrapped optional”. That is, it’s an optional, much like String?, but one that pretends to be a non-optional sometimes. Mostly for the purposes of reading it – you don’t have to explicitly unwrap it to get the value out. The downside being, if it is ever nil (and it totally can be), your code will trap and abort.
But this pretending-to-not-be-optional only goes so far. You cannot pass a String! to a function as inout when that argument is not an optional. Hence your problem.
Anton’s answer is completely correct in why it won’t work, and his suggested operator overload will make your code compile. But it isn’t the right solution – you should instead avoid using implicitly-unwrapped optionals as they are spring-loaded deathtraps and only to be used in specific circumstances (the most common being with Cocoa UI controls). 999 times out of 1,000 you would be better off with a regular optional or non-optional
Reason is that Foo, Foo? and Foo! are different types in Swift.
There are certain operators pre-defined for you "out-of-the-box" which allow great deal of transparency between Foo and Foo!, but still these types are not the same.
When it comes to strings, operator
func += (inout left: String!, right: String)
... is simply not defined.
If you declare it like:
func += (inout left: String!, right: String) {
left = left + right
}
... then your code should compile the way you like it, that is:
oneString! += anyString