ios - Swift:类型必须实现协议(protocol)并且是给定类的子类

Question

<分区>

Answer 1

你可以这样做:

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
}

这定义了一个函数，该函数返回类型为“SomeType”和协议(protocol)为“SomeProtocol”的对象，并返回符合这些条件的对象。