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Variable parameters have been deprecated

Available from Swift 2.2

Paul Hudson      @twostraws

Another deprecation, but again with good reason: var parameters are deprecated because they offer only marginal usefulness, and are frequently confused with inout.

To give you an example, here is the printGreeting() function modified to use var:

func printGreeting(var name: String, repeat repeatCount: Int) {
    name = name.uppercaseString

    for _ in 0 ..< repeatCount {
        print(name)
    }
}

printGreeting("Taylor", repeat: 5)

The differences there are in the first two lines: name is now var name, and name gets converted to uppercase so that "TAYLOR" is printed out five times.

Without the var keyword, name would have been a constant and so the uppercaseString line would have failed.

The difference between var and inout is subtle: using var lets you modify a parameter inside the function, whereas inout causes your changes to persist even after the function ends.

As of Swift 2.2, var is deprecated, and it's slated for removal in Swift 3.0. If this is something you were using, just create a variable copy of the parameter inside the method, like this:

func printGreeting(name: String, repeat repeatCount: Int) {
    let upperName = name.uppercaseString

    for _ in 0 ..< repeatCount {
        print(upperName)
    }
}

printGreeting("Taylor", repeat: 5)
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