Swift：type必须实现协议并且是给定类的子类 [英] Swift: type must implement protocol and be a subclass of given class

问题描述

在Objective-C中，您可以将类型定义为给定类的类并实现协议：

In Objective-C, you could define a type as being of a given class and implementing a protocol:

- (UIView <Protocol> *)someMethod;

这将告诉 someMethod 是 UIView 实现给定的协议协议。有没有办法在Swift中强制执行类似的东西？

This would tell that the value returned by someMethod was a UIView implementing a given protocol Protocol. Is there a way to enforce something similar in Swift?

推荐答案

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。

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

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