Swift:type必须实现协议并且是给定类的子类

Question

在Objective-C中,您可以将类型定义为给定类的类并实现协议:

- (UIView <Protocol> *)someMethod;

这将告诉返回的值someMethod是UIView实现给定协议Protocol.有没有办法在Swift中强制执行类似的操作？

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,它返回一个'SomeType'类型的对象和协议'SomeProtocol',并返回一个符合这些条件的对象.