Swift enum get an associated value without passing through a cases [duplicate] - swift

Question

I've got an enumeration with a few different cases which are different types, e.g.
enum X {
case AsInt(Int)
case AsDouble(Double)
}
I can switch on these just fine to get the underlying value back out. However, the switch statement is highly annoying with trying to make me execute some code for the other cases that I simply don't care about. For example, right now I have something like
func AsInt(x: X) -> Int? {
switch x {
case AsInt(let num):
return num;
default:
return nil;
}
}
This works but it's pretty tedious always having to reference this method and having to write a new one for each case of each enumeration. What I'm looking for is how to simply attempt to cast a value of type X to one of the cases, like
var object: X = func();
let value = obj as? Int;
if value {
// do shit
}
How can I simply check for a case without having to enumerate all of the cases about which I don't care and execute some non-statement for them?
Bonus points for any solution that can declare value as part of the conditional instead of polluting the scope.

Answer 1

There are actually multiple ways to do it.
Let's do it by extending your enum with a computed property:
enum X {
case asInt(Int)
case asDouble(Double)
var asInt: Int? {
// ... see below
}
}
Solutions with if case
By having let outside:
var asInt: Int? {
if case let .asInt(value) = self {
return value
}
return nil
}
By having let inside:
var asInt: Int? {
if case .asInt(let value) = self {
return value
}
return nil
}
Solutions with guard case
By having let outside:
var asInt: Int? {
guard case let .asInt(value) = self else {
return nil
}
return value
}
By having let inside:
var asInt: Int? {
guard case .asInt(let value) = self else {
return nil
}
return value
}
The last one is my personal favorite syntax of the four solutions.

Answer 2

As of Swift 2 (Xcode 7) this is possible with if/case and
pattern matching:
let x : X = ...
if case let .AsInt(num) = x {
print(num)
}
The scope of num is restricted to the if-statement.