Swift: type must implement protocol and be a subclass of given class [duplicate]

Answer 1

You can do it like this:

protocol SomeProtocol {
  func someMethodInSomeProtocol()
}

class SomeType { }

class SomeOtherType: SomeType, SomeProtocol {
  func someMethodInSomeProtocol() { }
}

class SomeOtherOtherType: SomeType, SomeProtocol {
  func someMethodInSomeProtocol() { }
}

func someMethod<T: SomeType where T: SomeProtocol>(condition: Bool) -> T {
  var someVar : T
  if (condition) {
    someVar = SomeOtherType() as T
  }
  else {
    someVar = SomeOtherOtherType() as T
  }

  someVar.someMethodInSomeProtocol()
  return someVar as T
}

This defines a function that returns an object of type 'SomeType' and protocol 'SomeProtocol' and returns an object that adheres to those conditions.