Alright so I just update Xcode to 7.3 and now I get this warning:
'var' parameters are deprecated and will be removed in Swift 3
How to fix this when I need to use the var in this function:
public func getQuestionList(var language: String) -> NSArray {
if self.data.count > 0 {
if (language.isEmpty) {
language = "NL"
}
return self.data.objectForKey("questionList" + language) as! NSArray
}
return NSArray()
}
Have you tried to assign to a new var
public func getQuestionList(language: String) -> NSArray {
var lang = language
if self.data.count > 0 {
if (lang.isEmpty) {
lang = "NL"
}
return self.data.objectForKey("questionList" + lang) as! NSArray
}
return NSArray()
}
var parameters for this reason. your solution is the best approach since it doesn't require introducing a new variable name. –
Lighterman The discussion of the removal of Var from a function parameter is fully documented within this submission on GitHub: Remove Var Parameters
In that document you will find that people often confuse var parameters with inout parameters. A var parameter simply means that the parameter is mutable within the context of the function, while with an inout parameter the value of the parameter at the point of return will be copied out of the function and into the caller's context.
The correct way to solve this problem is to remove var from the parameter and introduce a local var variable. At the top of the routine copy the parameter's value into that variable.
var name = name, just to be able to mutate the parameter inside the function. Something that would happen quite often I think. –
Scevor scala allows var parameters –
Aurie Have you tried to assign to a new var
public func getQuestionList(language: String) -> NSArray {
var lang = language
if self.data.count > 0 {
if (lang.isEmpty) {
lang = "NL"
}
return self.data.objectForKey("questionList" + lang) as! NSArray
}
return NSArray()
}
var parameters for this reason. your solution is the best approach since it doesn't require introducing a new variable name. –
Lighterman Just add this one line at the beginning of the function:
var language = language
and the rest of your code can stay unchanged, like this:
public func getQuestionList(language: String) -> NSArray {
var language = language
if self.data.count > 0 {
if (language.isEmpty) {
language = "NL"
}
return self.data.objectForKey("questionList" + language) as! NSArray
}
return NSArray()
}
if let and guard let statements. –
Delate A lot of people are suggesting an inout parameter, but that's really not what they're designed for. Besides, it doesn't allow calling the function with a let constant, nor with a string literal. Why don't you simply add the default value to the function signature?
public func getQuestionList(language language: String = "NL") -> NSArray {
if data.count > 0 {
return data.objectForKey("questionList" + language) as! NSArray
} else {
return NSArray()
}
}
Just make sure to not call getQuestionList with the empty string in case you want the default language, but just leave out the parameter:
let list = getQuestionList() // uses the default "NL" language
public func getQuestionList(language: inout String) -> NSArray {
if self.data.count > 0 {
if (language.isEmpty) {
language = "NL"
}
return self.data.objectForKey("questionList" + language) as! NSArray
}
return NSArray()
}
I think @Harris and @garanda answers are the best approach.
Anyway in your case, there isn't need of a var, you can do:
public func getQuestionList(language: String) -> NSArray {
if self.data.count > 0 {
return self.data.objectForKey("questionList" + (language.isEmpty ? "NL" : language)) as! NSArray
}
return NSArray()
}
In-Out Parameters
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. If you want a function to modify a parameter’s value, and you want those changes to persist after the function call has ended, define that parameter as an in-out parameter instead.
You write an in-out parameter by placing the inout keyword right before a parameter’s type. An in-out parameter has a value that is passed in to the function, is modified by the function, and is passed back out of the function to replace the original value. For a detailed discussion of the behavior of in-out parameters and associated compiler optimizations, see In-Out Parameters.
You can only pass a variable as the argument for an in-out parameter. You cannot pass a constant or a literal value as the argument, because constants and literals cannot be modified. You place an ampersand (&) directly before a variable’s name when you pass it as an argument to an in-out parameter, to indicate that it can be modified by the function.
NOTE
In-out parameters cannot have default values, and variadic parameters cannot be marked as inout.
Here’s an example of a function called swapTwoInts(::), which has two in-out integer parameters called a and b:
func swapTwoInts(_ a: inout Int, _ b: inout Int) {
let temporaryA = a
a = b
b = temporaryA
}
The swapTwoInts(::) function simply swaps the value of b into a, and the value of a into b. The function performs this swap by storing the value of a in a temporary constant called temporaryA, assigning the value of b to a, and then assigning temporaryA to b.
You can call the swapTwoInts(::) function with two variables of type Int to swap their values. Note that the names of someInt and anotherInt are prefixed with an ampersand when they are passed to the swapTwoInts(::) function:
var someInt = 3
var anotherInt = 107
swapTwoInts(&someInt, &anotherInt)
print("someInt is now \(someInt), and anotherInt is now \(anotherInt)")
// Prints "someInt is now 107, and anotherInt is now 3"
The example above shows that the original values of someInt and anotherInt are modified by the swapTwoInts(::) function, even though they were originally defined outside of the function.
NOTE
In-out parameters are not the same as returning a value from a function. The swapTwoInts example above does not define a return type or return a value, but it still modifies the values of someInt and anotherInt. In-out parameters are an alternative way for a function to have an effect outside of the scope of its function body.
Here is another idea. My use case was to pass around a string array to append to it, for which the array must be passed in mutably. I did not want to have state in my class for this either. So I made a class that holds the array and pass that. Depending on your use case it may seem silly to have a class that holds just that one variable.
private class StringBuilder {
var buffer: [String] = []
func append(_ str: String) {
buffer.append(str)
}
func toString() -> String {
return buffer.joined()
}
}
I only use append and joined methods on the array so it was easy to change the type with minimal other changes to my code.
Some example usage:
private func writeMap(map: LevelMap, url: URL) -> Bool {
let buffer = StringBuilder()
if !writeHeader(map: map, buffer: buffer) {
return false
}
if !writeFloors(map: map, buffer: buffer) {
return false
}
let content = buffer.toString()
do {
try content.write(to: url, atomically: true, encoding: .utf8)
return true
} catch {}
return false
}
private func writeHeader(map: LevelMap, buffer: StringBuilder) -> Bool {
buffer.append("something here ...\n")
return true
}
I think you should inout instead, following the recommendations given in this discussion.
To put it in context, in your example:
public func getQuestionList(var language: String)
should be replaced by:
public func getQuestionList(language: inout String)
© 2022 - 2024 — McMap. All rights reserved.
public func getQuestionList(inout language: String) -> NSArray– HairwormgetQuestionto have any side effects. – Delatefunc foo(_ bar: int) { /*use bar*/ bar+=1; foo(bar); }. This is impossible without var params. You either need to create a separate var within the function and copy the value, or mark the param as inout. The former is slow, the latter causes undefined behaviour. Many algorithms use recursion like this. – Leivafoo(bar + 1)? – Albersbaris not an int but a complex object? – Leiva