Access enum associated value as optional

Question

How can I access an enum value for a specific case without having to implement an enum function for each case?

I\'m thinking something like this:

enum Re         


        

Answer 1

Add two computed properties for success case and failure case respectively.

enum Result<S, F> {
  case success(S)
  case failure(F)

  var successValue: S? {
    switch self {
    case .success(let value):
      return value
    case .failure:
      return nil
    }
  }

  var failureValue: F? {
    switch self {
    case .failure(let value):
      return value
    case .success:
      return nil
    }
  }
}


let resultSuccess = Result<Int, String>.success(1)
let resultFailure = Result<Int, String>.failure("Error")

if let error = resultFailure.failureValue {
  // do something
  print(error)
}

Answer 2

I'm afraid this is the most you can get:

    let error: String?
    if case .failure(let e) = resultFailure {error = e}

Answer 3

This technique from the question "Accessing an Enumeration association value in Swift" will do it:

enum Result<S, F> {
  case success(S)
  case failure(F)
}

let resultFailure = Result<Int, String>.failure("Error")

var error: String?

if case let Result.failure(value) = resultFailure {
  error = value
}

print(error) // Optional("Error")

This will also work in place of the "if case let =" construct:

switch resultFailure {
case let Result.failure(value): error = value
default                       : error = nil
}

Here you go, 1 line and uses let:

let error: String? = { if case let .failure(value) = resultFailure { return value }; return nil }()