I am trying to write in swift something that should be very basic, but I can't seem to get a handle on it :
First, I create a global variable. For example:
var xx:Int
Then, I want to create a conditional instruction. Something like :
if (xx == 1){
//do something
}
else if (xx == 2) {
//do something else
}
I can do this very easily in Objective-C, but I can't seem to be able to do it in Swift. I have looked everywhere, and don't seem to find the answer.
With the code you provided you're probably getting the error: "Variable xx used before initialized". This is happening because the declaration of the variable is incomplete, you neither gave a value to the variable nor told the compiler it is an optional. You have three options:
Give a initial value to it; var xx: Int = //value here
Declare it as an optional (doing this you say that it may not have a value, if it does the code will be executed, if it doesn't it won't); var xx: Int?
Force unwrap the variable (it still an optional, but if you force-unwrap it you're assuring the compiler that the variable will have a value when needed, otherwise it'll crash); var xx: Int!
Or you can say var xx = Int() that way it's initialized and the default initialization is equal to 0. This is different than the other answers and allows you to have a value from the get go if you're not sure what value might be assigned during runtime.
In addition to the other poster's point that you must assign an initial value before you can use xx, you also need to lose the parentheses around the condition in your if statement:
var xx:Int
xx = 2
if xx == 1
{
//do something
}
else if xx == 2
{
//do something else
}
Related
What is the mechanism of declaring w/o value in Swift5 ? Does the first assign become the real declaration ?
And, should we avoid to declare without value in Swift?
var v:String;
if true {
v = "Hello"
print(v) // print "Hello" when without the print below
}
print(v) // variable 'v' used before being initialized
var v:String="";
if true {
v = "Hello"
print(v) // print "Hello"
}
print(v) // print "Hello"
Well, the message is not very helpful, and that's the problem. This pattern (which I call computed initialization) is perfectly legal and useful and — amazingly — you can even use let instead of var. But you must initialize the uninitialized variable by all possible paths before you use it. So you have:
var v:String
if true {
v = "Hello"
}
print(v) // error
But hold my beer and watch this:
var v:String
if true {
v = "Hello"
} else {
v = "Goodbye"
}
print(v) // fine!
Or even:
let v:String
if true {
v = "Hello"
} else {
v = "Goodbye"
}
print(v) // fine!
Amazing, eh?
Now, you might say: OK, but true will always be true so it's silly to make me fulfill the "all paths" rule. Too bad! The compiler insists anyway, and then lets you off later with a warning that the else won't be executed. But a warning lets you compile; an error doesn't. The truth is that your example is very artificial. But this is a real-life possibility:
let v:String
if self.someBoolProperty {
v = "Hello"
} else {
v = "Goodbye"
}
print(v) // fine!
Not only is this sort of thing legal, it is actually the pattern that Apple recommends under certain slightly tricky circumstances. For instance, it is used in Apple's own example code showing how to use the Swift 5 Result struct:
let result: Result<Int, EntropyError>
if count < AsyncRandomGenerator.entropyLimit {
// Produce numbers until reaching the entropy limit.
result = .success(Int.random(in: 1...100))
} else {
// Supply a failure reason when the caller hits the limit.
result = .failure(.entropyDepleted)
}
So this is because swift compiles your code and notices that your value var v:String; is undeclared before being used which makes it an "optional" value. Even though you are assigning it within the if statement, if you were to get rid of the true value it is possible that the if statement would never run therefore no value will ever be stored in v, thus it would be used before "assigned".
So to answer your question if you want your value to be an optional and possible empty value declare v as the following var v:String? if you would like it to be a non-optional value with a value always stored within v you should declare it as the following var v = "". Swift will interpret this declaration as a String.
To answer your second question, defining without values in swift is 100% allowed, it really just depends on how you want to handle your code. I use optional values in my code all the time, but I don't necessarily like optionals so i usually like to declare my values such as var v = "" that way if the value is empty my UI or whatever else i'm manipulating won't break. But if i need to ensure a value is present i will have to make my value optional so i can use an if statement to check whether it's a valid value or not.
Shorter version of what I'm trying to say is, you are receiving the compiler warning because you are declaring v as a non-optional value rather than an optional value.
I know there is another thread with the same question, but it doesn't tell what is actually causing the problem
Im new to swift, so Im a bit confused on this.
I wrote a very simple program that is supposed to start with a default number of followers (0) and assign that to 'defaultfollowers' and once that becomes 1 its supposed become "followers", but I get the error "Type 'Int' does not conform to protocol 'BooleanType'". What is causing this and why
var followerdeafault = 0
var followers = 0
if (followerdeafault++){
var followers = followerdeafault
}
In Swift you can't implicitly substitute Int instead of Bool. This was done to prevent confusion and make code more readable.
So instead of this
let x = 10
if x { /* do something */ }
You have to write this:
let x = 10
if x != 0 { /* do something */ }
Also you can't pass an Optional instead of Bool to check if it's nil, as you would do in Objective-C. Use explicit comparison instead:
if myObject != nil { /* do something */ }
As the comments said, you're trying to use an Int in a Bool comparison statement. What you're looking for is probably something like this:
if followerdeafuaut++ == 1 { ... }
Also side note: the ++ operator is deprecated, moving towards using +=
I'm very new to swift, but proficient in other languages like Java, JavaScript, C, ... I'm lost with Swift syntax when it comes to create expressions. Look at this basic example where I just try to find out if one string is contained into another by calling String.rangeOfString that returns an Optional Range (Range?)
This works as expected:
let LEXEMA:String="http://"
let longUrl:String="http://badgirls.many/picture.png"
let range=longUrl.rangeOfString(LEXEMA);
if (range? != nil) {
// blah
}
Now I'm trying to combine the expression inside the if, something like:
if (longUrl.rangeOfString(LEXEMA)? !=nil) {
// blah
}
But I always get syntax errors, the above yields a "Expected Separator" and can't understand why. Done some more tests:
if (absolutePath.rangeOfString(URL_LEXEMA) !=nil) { }
Expected Separator before "!"
if absolutePath.rangeOfString(URL_LEXEMA) !=nil { }
Braced block of statements is an unused closure
What am I doing wrong?
If you’re coming from other like Java, you might be thinking of optionals like pointers/references, and so used to equating them to nil and if non-nil, using them. But this is probably going to lead to more confusion. Instead, think of them like a container for a possible result, that you need to unwrap to use. if let combines the test and unwrapping operation.
With this in mind, here’s how you could adapt your code:
let LEXEMA: String="http://"
let longUrl: String="http://badgirls.many/picture.png"
if let range = longUrl.rangeOfString(LEXEMA) {
// use range, which will be the unwrapped non-optional range
}
else {
// no such range, perhaps log an error if this shouldn’t happen
}
Note, that ? suffixing behaviour you were using changes in Swift 1.2 so even the code in your question that compiles in 1.1 won’t in 1.2.
It’s possible that sometimes you are whether there was a value returned, but you don’t actually need that value, just to know it wasn’t nil. In that case, you can compare the value to nil without the let:
if longUrl.rangeOfString(LEXEMA) != nil {
// there was a value, but you don't care what that value was
}
That said, the above is probably better expressed as:
if longUrl.hasPrefix(LEXEMA) { }
For starters:
You don't need parenthesis with if statements unless you have nested parenthetical subexpressions that require it.
You don't need to specify the type on the left side of the = of a let or var declaration if Swift can figure it out from the right side of the =. Very often Swift can figure it out, and you can tell that Swift can figure it out, so you can avoid that redundant clutter.
You do need to specify the type if Swift cannot figure out the type from
the right side. Example:
For example, consider the following lines:
let LEXEMA = "http://"
let longUrl = "http://badgirls.many/picture.png"
Swift can figure out that they're strings.
Similarly for this function or class that returns a UIView:
var myView = ViewReturningClassOrFunc()
Consider this:
#IBOutlet var myView : UIView!
In the above line, Swift cannot figure out ahead of time it will be assigned a UIView, so you have to provide the type. By providing a ! at the end you've made it an implicitly unwrapped optional. That means, like ?, you're indicating that it can be nil, but that you are confident it will never be nil at the time you access it, so Swift won't require you to put a ! after it when you reference it. That trick is a time saver and big convenience.
You should NOT add the ? to the line:
if (longUrl.rangeOfString(URL_LEXEMA) !=nil) {
As another answer pointed out, you're missing the let.
if let longUrl.rangeOfString(URL_LEXEMA) {
println("What do I win? :-)")
}
swift is case sensitive language. you need to check about whitespaces as well
if longUrl.rangeOfString(LEXEMA) != nil {
//your condition
}
there should be space between statement != nil
Just add a space between != and nil like:
if longUrl.rangeOfString(LEXEMA) != nil {
// blah
}
I tested your code in playground, an error of Expected ',' separator reported.
And do not forget the rules that 1s and 0s and Airspeed Velocity said.
Why do assignments involving Swift optionals type check?
For example in,
var foo : Int? = 0
foo = foo!
foo and foo! do not have the same type. Shouldn't you need to wrap the unwrapped value to assign it to an optional type?
This is part of the syntactic sugar behind optionals. Assigning a non-optional value is how you wrap it in the optional type.
Since an optional indicates the presence or absence of a value, you shouldn't have to do anything special to indicate the presence of a value other than provide one. For example, in a function:
func gimmeSomethingMaybe() -> String? {
if arc4random_uniform(10) > 7 {
return "something"
} else {
return nil
}
}
Imagine if every time you wanted to return a real value from a function that's capable of returning nil, you had to write return Optional(value). That'd get old pretty fast, right? Optionals are an important feature of the language — even though they're actually implemented by the standard library, the syntactic sugar / automatic wrapping is there to keep it from being tedious to use them.
Edit: just to go a bit further into this... the sugar also helps to enforce the notion that a real value should not be optional. For example:
let one = 1
one? // error (in Swift 1.2, allowed but meaningless in Swift 1.1)
"two"? // error (ditto)
You can create an optional wrapping a real value with the Optional(one) initializer, but that has little semantic meaning on its own, so you almost never need to.
Optionals should come into play when there's "mystery" as to whether a value is present or absent — that is, when whether one part of a program receives a value (or no value) depends on state unknown to that part of the program. If you know you have a real value, there's no mystery... instead, you let the unknown come into play at the boundary between the code that knows the value and the code that doesn't know — that is, the function/method/property definition that hands that value off to somewhere.
After reading rickster's answer, I came up with a simple laymen terms answer. To me the whole whole gist of his answer is
Since an optional indicates the presence or absence of a value, you
shouldn't have to do anything special to indicate the presence of a
value other than provide one
An optional is an enum. Which has 2 cases a or b.
String?
|
An enum with 2 cases
|
a b
| |
Set notSet
| |
any value like "hi" nil
So you could do either of the things when you want to assign to an optional.
Say the value is either:
a: it's set to "hi"
b: it's not set, so it's nil
c: it just equals to another enum ie another optional
code:
var str : String?
var anotherOptional : String?
str = nil // nil <-- this is like case b
str = "hi" // "hi" <-- this is like case a
str = anotherOptional // nil <-- this is like case c
anotherOptional = "hi again"
str = anotherOptional // "hi again" <-- this is like case c
I'm learning Swift and I'm reading The Swift Programming Language from Apple. I don't have any Objective-C background (only PHP, JavaScript, and others, but not Objective-C).
On page 24-25 I see this code:
//...Class definition stuff...
var perimeter: Double {
get {
return 3.0 * sideLength
}
set {
sideLength = newValue / 3.0
}
}
//...Class continues...
This part is not specified in the book, and I can't get what those are for.
What are get and set?
The getting and setting of variables within classes refers to either retrieving ("getting") or altering ("setting") their contents.
Consider a variable members of a class family. Naturally, this variable would need to be an integer, since a family can never consist of two point something people.
So you would probably go ahead by defining the members variable like this:
class family {
var members: Int
}
This, however, will give people using this class the possibility to set the number of family members to something like 0 or 1. And since there is no such thing as a family of 1 or 0, this is quite unfortunate.
This is where the getters and setters come in. This way you can decide for yourself how variables can be altered and what values they can receive, as well as deciding what content they return.
Returning to our family class, let's make sure nobody can set the members value to anything less than 2:
class family {
var _members: Int = 2
var members: Int {
get {
return _members
}
set (newVal) {
if newVal >= 2 {
_members = newVal
} else {
println('error: cannot have family with less than 2 members')
}
}
}
}
Now we can access the members variable as before, by typing instanceOfFamily.members, and thanks to the setter function, we can also set it's value as before, by typing, for example: instanceOfFamily.members = 3. What has changed, however, is the fact that we cannot set this variable to anything smaller than 2 anymore.
Note the introduction of the _members variable, which is the actual variable to store the value that we set through the members setter function. The original members has now become a computed property, meaning that it only acts as an interface to deal with our actual variable.
A simple question should be followed by a short, simple and clear answer.
When we are getting a value of the property it fires its get{} part.
When we are setting a value to the property it fires its set{} part.
PS. When setting a value to the property, Swift automatically creates a constant named "newValue" = a value we are setting. After a constant "newValue" becomes accessible in the property's set{} part.
Example:
var A:Int = 0
var B:Int = 0
var C:Int {
get {return 1}
set {print("Recived new value", newValue, " and stored into 'B' ")
B = newValue
}
}
// When we are getting a value of C it fires get{} part of C property
A = C
A // Now A = 1
// When we are setting a value to C it fires set{} part of C property
C = 2
B // Now B = 2
You should look at Computed Properties.
In your code sample, perimeter is a property not backed up by a class variable. Instead its value is computed using the get method and stored via the set method - usually referred to as getter and setter.
When you use that property like this:
var cp = myClass.perimeter
you are invoking the code contained in the get code block, and when you use it like this:
myClass.perimeter = 5.0
You are invoking the code contained in the set code block, where newValue is automatically filled with the value provided at the right of the assignment operator.
Computed properties can be read/write if both a getter and a setter are specified, or read-only if the getter only is specified.
A variable declares and is called like this in a class:
class X {
var x: Int = 3
}
var y = X()
print("value of x is: ", y.x)
//value of x is: 3
Now you want the program to make the default value of x more than or equal to 3. Now take the hypothetical case if x is less than 3: your program will fail.
So, you want people to either put in 3 or more than 3. Swift got it easy for you and it is important to understand this bit - it is an advanced way of dating the variable value, because they will extensively use it in iOS development. Now let's see how get and set will be used here.
class X {
var _x: Int = 3
var x: Int {
get {
return _x
}
set(newVal) { // Set always take one argument
if newVal >= 3 {
_x = newVal // Updating _x with the input value by the user
print("new value is: ", _x)
}
else {
print("error must be greater than 3")
}
}
}
}
let y = X()
y.x = 1
print(y.x) // The error must be greater than 3
y.x = 8 // // The new value is: 8
If you still have doubts, just remember the use of get and set is to update any variable the way we want it to be updated. get and set will give you better control to rule your logic. It is a powerful tool, hence not easily understandable.