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)
SPONSORED Spend less time managing in-app purchase infrastructure so you can focus on building your app. RevenueCat gives everything you need to easily implement, manage, and analyze in-app purchases and subscriptions without managing servers or writing backend code.
Sponsor Hacking with Swift and reach the world's largest Swift community!
Other changes in Swift 2.2…
- ++ and -- are deprecated
- Traditional C-style for loops are deprecated
- Comparing tuples
- Tuple splat syntax is deprecated
- More keywords can be used as argument labels
- Renamed debug identifiers: line, function, file
- Stringified selectors are deprecated
- Compile-time Swift version checking
Download all Swift 2.2 changes as a playground Link to Swift 2.2 changes
