In Swift, how does a method return a generic type?

Question

I want a method to return String or Int as in the method below. How to define the * type? Mind, I dont want to return a tuple, just a single value. :beer:

func returnsSomething(key: Bool) -> * {
    if key == true {
        return "Im a string!"
    } else {
        return 42
    }
}
returnsSomething(true)

Answer 1

There are two options.

---

First, you can declare your method as returning type Any , which will allow you to return any type you want.

func returnsSomething(key: Bool) -> Any {
    if key {
        return "I'm a string"
    } else {
        return 42
    }
}

This will work perfectly fine, but it has the downside that if you use implicit type definitions:

let foo = returnsSomething(true)

Then foo won't have the type of String or Int , but rather Any , and you're left with optional downcasting.

let foo = returnsSomething(true)
if let bar = foo as? Int {
    // use bar as an Int
} else if let bar = foo as? String {
    // use bar as a String
}

^{As a note, you don't actually have to use Any here. For example, if you wanted to return a UIView , UIButton , or UITableView , you could use UIView , as the latter two are both subclasses of UIView .}

---

Second, you can use method overloading:

func returnsSomething() -> String {
    return "I'm a string!"
}

func returnsSomething() -> Int {
    return 42
}

This works, but now you cannot use implicit typing at all with this method. Notice that the following generates an error:

 let foo = returnsSomething() 

Ambiguous use of 'returnsSomething'

However, if we're explicit with the type, this is allowed:

let fooString: String = returnsSomething()
let fooInt: Int = returnsSomething()

And the compiler is perfectly satisfied.