在 XPC 协议中使用自定义 class

Using custom class in XPC protocol

我正在尝试在 withReply 签名上使用我自己的类型编写 XPC 服务。 type/class 拥有主应用程序和 XPC 服务的 Xcode 的“目标成员资格”。但是,即使在 withReply 签名中使用了相同的 class,但我在调试输出中得到了 incompatible reply block signature,但是 Xcode 目标不同,我将在下面解释。

注意:这是在 Swift 中完成的,使用 this project 让我开始。除了那里他们使用 NSData 而不是自定义类型。

详情

为了这个问题的目的,我将使用下面的例子

然后是所有常用的 XPC 样板文件,我在其中导出符合 TweetTransfer 的对象。 XPC 服务似乎已启动,但随后在它与主应用程序之间传输失败并显示

XPCWorker[11069:402719] <NSXPCConnection: 0x61800010e220> connection from pid 11066 received an undecodable message

下面是完整的消息[1],但“wire”和“local”之间的唯一区别是参数一是

其中 Xcode 目标不同。这足以把它扔掉吗?那么如何在应用程序和它的 XPC 服务之间共享代码?

[1] 完整错误文本

<NSXPCConnection: 0x61800010e220> connection from pid 11066 received an undecodable message (incompatible reply block signature (wire: <NSMethodSignature: 0x618000074ec0>
    number of arguments = 2
    frame size = 224
    is special struct return? NO
    return value: -------- -------- -------- --------
        type encoding (v) 'v'
        flags {}
        modifiers {}
        frame {offset = 0, offset adjust = 0, size = 0, size adjust = 0}
        memory {offset = 0, size = 0}
    argument 0: -------- -------- -------- --------
        type encoding (@) '@?'
        flags {isObject, isBlock}
        modifiers {}
        frame {offset = 0, offset adjust = 0, size = 8, size adjust = 0}
        memory {offset = 0, size = 8}
    argument 1: -------- -------- -------- --------
        type encoding (@) '@"_TtC17MainApp5Tweet"'
        flags {isObject}
        modifiers {}
        frame {offset = 8, offset adjust = 0, size = 8, size adjust = 0}
        memory {offset = 0, size = 8}
            class '_TtC17MainApp5Tweet'
 vs local: <NSMethodSignature: 0x610000074740>
    number of arguments = 2
    frame size = 224
    is special struct return? NO
    return value: -------- -------- -------- --------
        type encoding (v) 'v'
        flags {}
        modifiers {}
        frame {offset = 0, offset adjust = 0, size = 0, size adjust = 0}
        memory {offset = 0, size = 0}
    argument 0: -------- -------- -------- --------
        type encoding (@) '@?'
        flags {isObject, isBlock}
        modifiers {}
        frame {offset = 0, offset adjust = 0, size = 8, size adjust = 0}
        memory {offset = 0, size = 8}
    argument 1: -------- -------- -------- --------
        type encoding (@) '@"_TtC23XPCWorker5Tweet"'
        flags {isObject}
        modifiers {}
        frame {offset = 8, offset adjust = 0, size = 8, size adjust = 0}
        memory {offset = 0, size = 8}
            class '_TtC23XPCWorker5Tweet'
)

更新

关于协议、remoteObjectProxy 连接和 Tweet 对象的更多信息。这是用于 XPC 调用的协议:

@objc(TweetTransfer)
protocol TweetTransfer {
  func take(_ count: Int, withReply: replyType)
}

typealias replyType = ((Tweet) -> Void)

为了方便起见,我使用了类型别名。然后 Tweet 对象非常简单,仅用于测试(尽管支持 NSSecureCoding 有点复杂):

final class Tweet: NSObject, NSSecureCoding {
  var name: String
  var text: String
  static var supportsSecureCoding = true

  init(name: String, text: String) {
    self.name = name
    self.text = text
  }

  init?(coder aDecoder: NSCoder) {
    guard let name = aDecoder.decodeObject(forKey: "name") as? String else {
      fatalError("Could not deserialise name!")
    }

    guard let text = aDecoder.decodeObject(forKey: "text") as? String else {
      fatalError("Could not deseralise text!")
    }

    self.name = name
    self.text = text
  }

  func encode(with aCoder: NSCoder) {
    aCoder.encode(name, forKey: "name")
    aCoder.encode(text, forKey: "text")
  }
}

最后是我们调用 remoteObjectProxy 的点

guard let loader = workerConnection.remoteObjectProxyWithErrorHandler(handler) as? TweetTransfer else {
  fatalError("Could not map worker to TweetTransfer protocol!")
}

var tweets = [Tweet]()
loader.take(1) { tweet in
  tweets.append(tweet)
}

The full message is below but the only difference between the “wire” and “local” is that argument one is

  • wire - _TtC17MainApp5Tweet
  • local - _TtC23XPCWorker5Tweet

Where the Xcode target is different. Is that enough to throw it off? How then do I share code between an app and it's XPC service?

这确实足以摆脱它。 Swift 的命名空间使归档对象显示为不同的 class。您可以通过声明您的 Tweet 对象来禁用名称间距;

@objc(Tweet) class Tweet: NSObject, NSSecureCoding { ... }

@objc(name) 中的 name 通常作为在 objc 与 Swift 中呈现不同名称的一种方式呈现,但它也具有以下效果禁用 Swift 的名称间距。

来自Using Swift with Cocoa and Objective-C

When you use the @objc(name) attribute on a Swift class, the class is made available in Objective-C without any namespacing. As a result, this attribute can also be useful when migrating an archivable Objective-C class to Swift. Because archived objects store the name of their class in the archive, you should use the @objc(name) attribute to specify the same name as your Objective-C class so that older archives can be unarchived by your new Swift class.

另一种选择是将您的自定义对象移至框架。该框架目标随后成为命名空间,XPC 和 App 将在框架中引用相同的 namespace/class。