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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