I'm just going through some Swift tuts that are obviously already outdated as of Beta3 ...
func exchange<T>(data:[T], i:Int, j:Int)
{
let temp = data[i];
data[i] = data[j];
data[j] = temp;
}
Playgrounds tells me:
Error: @lvalue $T8 is not identical to T.
How do I change it to make it work?
Arrays in Swift are value types. That means that data is copied when passed into your exchange method, but you are trying to modify the copy to affect the original version. Instead you should do one of two things:
data as an inout parameter:func exchange<T>(inout data:[T], i:Int, j:Int)
Then when calling it you have to add an & before the call:
var myArray = ["first", "second"]
exchange(&myArray, 0, 1)
func exchange<T>(data:[T], i:Int, j:Int) -> [T]
{
var newData = data
newData[i] = data[j]
newData[j] = data[i]
return newData
}
I recommend this way over the in-out parameter because in-out parameters create more complicated state. You have two variables pointing to and potentially manipulating the same piece of memory. What if exchange decided to do its work on a separate thread? There is also a reason that Apple decided to make arrays value types, using in-out subverts that. Finally, returning a copy is much closer to Functional Programming which is a promising direction that Swift can move. The less state we have in our apps, the fewer bugs we will create (in general).
inout parameters? that is more efficient in the case of huge arrays than making a copy of the array every time when the method is called...
inout parameters are totally reliable! the only weak point is the developer, but if the developer is under-educated (that leads a poor implementation which leads having more bugs which leads spending more time on fixing bugs, etc...), does not matter which version they use, they will make logical-structural mistakes eventually in a project. however, I guess we all have different debug policies, personally I spend 10-15% of the development time fixing bugs, but I've seen 'developers' who spent 60-65% – and you know what that means.inout it as a valid solution for a reason.