I am trying some simple code, which I found on natashatherobot.com.
var str = "Hello, playground"
let rangeOfHello = Range(start: str.startIndex, end: advance(str.startIndex, 5))
let helloStr = str.substringWithRange(rangeOfHello)
return helloStr
It works fine when I try it in Playgrounds:
But when I try using it in my Xcode project it gives me a compilation error:
Any ideas why this is happening?
In your function declaration you are saying that it returns a void, but you are trying to return a string, You need to add the -> String in the end of your function to match what you are trying to do
your function:
func getStringBetween(startString: String, endString: String) -> ()
shoud be:
func getStringBetween(startString: String, endString: String) -> String
You did not specify the return type of your func:
func getStringBetween(startString: String, endString: String) -> String {
The problem is the method getStringBetween(:endString:), which you defined in your extension for String.
This method does not define a return type (there's no -> Type after the parameter list). Therefore the implicit return type is Void, so your method could be written as:
func getStringBetween(startString: String, endString: String) -> Void
In Swift the type Void is equivalent to an empty tuple ().
The problem is therefore that Swift expects a return type of Void denoted by (), but you're returning a String.
This problem is easily fixed by adding the return type String to the declaration of your method:
func getStringBetween(startString: String, endString: String) -> String
Side Note:
When writing return, without a specification of what is to be returned, Swift automatically returns ().