ios 选择器与动作 Swift 4
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Selector vs Action Swift 4
提问by user2511882
New to Swift. I have two snippets below:
Swift 新手。我有以下两个片段：
NotificationCenter.default.addObserver(self,
selector:#selector(ViewController.notificationReceived),
name: Notification.Name(rawValue: name), object: nil)
@objc func notificationReceived(notification:Notification){
let x = notification.userInfo!
print("\(x["name"]!)")
}
and finally
最后
let x:UITapGestureRecognizer = UITapGestureRecognizer(target: self,
action: #selector(tapped))
self.addGestureRecognizer(x)
func tapped(){
print("tapped")
self.delegate!.theViewTapped()
}
Why is it that for the notificationCenter? I am supposed to provide the @objctag for the selectorparameter but not for the UITapGestureRecognizeraction parameter?
为什么是这样notificationCenter？我应该@objc为selector参数提供标签而不是为UITapGestureRecognizer动作参数提供标签？
What exactly is the difference between Selectorand Action inSwift?
Selector和Action inSwift到底有什么区别？
回答by OOPer
Check this proposal for Swift 4: SE-0160 Limiting @objc inference
检查 Swift 4 的这个提议： SE-0160 限制 @objc 推理
According to the description in the proposal, your second code snippet also needs @objc.
根据提案中的描述，您的第二个代码段也需要@objc.
In fact, Swift 4 compiler bundled with Xcode 9 beta2 generates this error for the line using #selector(tapped):
实际上，与 Xcode 9 beta2 捆绑在一起的 Swift 4 编译器使用以下命令为该行生成此错误#selector(tapped)：
error: argument of '#selector' refers to instance method 'tapped()' that is not exposed to Objective-C
note: add '@objc' to expose this instance method to Objective-C
错误：“#selector”的参数引用了未暴露给 Objective-C 的实例方法“tapped()”
注意：添加“@objc”以将此实例方法公开给Objective-C
Maybe your second is a little bit too old to use with Swift 4. You better think all methods invoked through selector need @objcattribute.
也许你的第二个有点太老了，无法与 Swift 4 一起使用。你最好认为所有通过选择器调用的方法都需要@objc属性。
