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iOS Swift 2.0 Functions Function Parameters Function Scope

Michael Boswell
Michael Boswell
2,363 Points
Posted by Michael Boswell
Michael Boswell
2,363 Points

Swift 3 Local Variable Parameter

Similarly to increment and decrement, Xcode is telling me of more changes coming in Swift 3.

In this video, Pasan says that an argument by default is a constant, and to change that you put var in front of it. He explains it as thus:

Constant

func countDownAndUp(a: Int) {
  // do something
}

Variable

func countDownAndUp(var a: Int) {
  // do something
}

With Swift 3 deprecating var here, does that mean it's a var by default?

1 Answer

jcorum
jcorum
71,830 Points
jcorum
jcorum
71,830 Points

Michael, here's an excerpt from Swift.org (https://swift.org/blog/swift-2-2-new-features/):

var parameters are deprecated

Prior to Swift 2.2, function parameters could be declared as var if you wanted to modify them inside the function. For example:

func greet(var name: String) {
    name = name.uppercaseString
    print("Hello, \(name)!")
}

var name = "Taylor"
greet(name)
print("After function, name is \(name)")

While this was a helpful shortcut, it did add some extra confusion: does that final print() statement output “Taylor” or “TAYLOR”? This was made even more confusing by the presence of the inout keyword: using inout rather than var in that example, then adding a single ampersand, produces this code:

func greet(inout name: String) {
    name = name.uppercaseString
    print("Hello, \(name)!")
}

var name = "Taylor"
greet(&name)
print("After function, name is \(name)")

When run, the var example produces different output to the inout example, because changes to var parameters apply only inside the function whereas changes to inout parameters affect the original value directly.

In Swift 2.2, this confusion is cleared up by deprecating the var keyword for function parameters, ahead of its removal in Swift 3.0. If you want to replicate the old behavior, simply create your own copy inside the function like this:

func greet(name: String) {
    let uppercaseName = name.uppercaseString
    print("Hello, \(uppercaseName)!")
}

Looks like parameters will be constants unless marked inout.