将isKindOfClass与Swift一起使用

Question

我正在尝试学习Swift lang，我想知道如何将以下Objective-C转换为Swift：

- (void)touchesBegan:(NSSet *)touches withEvent:(UIEvent *)event {
    [super touchesBegan:touches withEvent:event];
    UITouch *touch = [touches anyObject];
    if ([touch.view isKindOfClass: UIPickerView.class]) {
      //your touch was in a uipickerview ... do whatever you have to do
    }
}

更具体地说，我需要知道如何在新语法中使用isKindOfClass。

override func touchesBegan(touches: NSSet, withEvent event: UIEvent) {
    ???
    if ??? {
        // your touch was in a uipickerview ...
    }
}

Answer 1

合适的Swift运算符是is:

if touch.view is UIPickerView {
    // touch.view is of type UIPickerView
}

当然，如果您还需要将视图分配给一个新的常量，那么if let ... as? ...语法就是您的孩子，正如Kevin提到的那样。但是，如果您不需要值，只需要检查类型，那么您应该使用is运算符。

Answer 2

override func touchesBegan(touches: NSSet, withEvent event: UIEvent) {
    super.touchesBegan(touches, withEvent: event)
    let touch : UITouch = touches.anyObject() as UITouch
    if touch.view.isKindOfClass(UIPickerView)
    {
    }
}

编辑

正如@Kevin的回答中所指出的，正确的方法是使用可选的类型转换运算符as?。您可以在Optional Chaining小节Downcasting小节中了解更多信息。

编辑2

正如用户@KPM在另一个答案中指出的那样，使用is运算符是正确的方法。

Answer 3

您可以将检查和强制转换组合为一个语句：

let touch = object.anyObject() as UITouch
if let picker = touch.view as? UIPickerView {
    ...
}

然后可以在if块中使用picker。

Answer 4

我会使用：

override func touchesBegan(touches: NSSet, withEvent event: UIEvent) {
    super.touchesBegan(touches, withEvent: event)
    let touch : UITouch = touches.anyObject() as UITouch
    if let touchView = touch.view as? UIPickerView
    {
    }
}

Answer 5

使用新Swift 2语法的另一种方法是使用guard并将其全部嵌套在一个条件中。

guard let touch = object.AnyObject() as? UITouch, let picker = touch.view as? UIPickerView else {
    return //Do Nothing
}
//Do something with picker