Pattern match and conditionally bind in a single Switch statement

Question

Is there a way to write this if/else if/else ladder as a switch statement?

let x: Any = "123"

if let s = x as? String {
    useString(s)
}
else if let i = x as? Int {
    useInt(i)
}
else if let b = x as? Bool {
    useBool(b)
}
else {
    fatalError()
}

Here's my attempt:

switch x {
case let s where s is String:   useString(s)
case let i where i is Int:      useInt(i)
case let b where b is Bool:     useBool(b)
default: fatalError()
}

It successfully chooses the right path, but s/i/b are still of type Any. The is check doesn't have any effect in casting them. This forces me to force cast with as! before usage.

Is there a way to switch on the type, and bind it to a name, all in one switch statement?

Habitforming

What you are doing is correct but missing executable statements for each case. Just add a print and see you wont see any errors – Homophone

Yes, that's an error, but I omitted it for the sake of demonstration. print(_:) has no problem handling parameters of type Any. Imagine the comments like //use s are actually function calls that take a parameter of type String/Int/Bool, respectively. – Habitforming

You said - "but s/i/b are still of type Any. The is check doesn't have any effect in casting them" So if you really print s/i/b then actually it prints the right value. print(s). I didn't knw about your function useString. Thanks – Homophone

Actually it is demonstrated as an example in the Swift Book: developer.apple.com/library/ios/documentation/Swift/Conceptual/…, search for for thing in things ... – Negligee

Answer 1

Sure, you can use the conditional casting pattern case let x as Type:

let x: Any = "123"

switch x {
case let s as String:
    print(s)   //use s
case let i as Int:
    print(i)   //use i
case let b as Bool:
    print(b)   //use b
default:
    fatalError()
}

Freer

Wonderful, this is exactly what I was looking for. I'll accept in 5 minutes – Habitforming

@AlexanderMomchliov Happy to help :) – Freer