将isKindOfClass与Swift一起使用

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将isKindOfClass与Swift一起使用

dsjdsjdsjjk_896 发布于 2023-01-11 19:54

我想尝试一下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isKindOfClass在新语法中使用.

override func touchesBegan(touches: NSSet, withEvent event: UIEvent) {

    ???

    if ??? {
        // your touch was in a uipickerview ...

    }
}

KPM.. 461

正确的Swift运算符是is:

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

当然,如果你还需要将视图分配给一个新的常量,那么if let ... as? ...语法就是你的男孩,正如凯文提到的那样.但是如果您不需要该值并且只需要检查类型,那么您应该使用is运算符.

3 个回答

正确的Swift运算符是is:
```
if touch.view is UIPickerView {
    // touch.view is of type UIPickerView
}
```
当然,如果你还需要将视图分配给一个新的常量,那么if let ... as? ...语法就是你的男孩,正如凯文提到的那样.但是如果您不需要该值并且只需要检查类型,那么您应该使用is运算符.
2023-01-11 19:54 回答

___周麗
```
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的回答所指出的那样,正确的方法是使用可选的类型转换运算符as?.您可以在section Optional Chaining子节中阅读更多相关信息Downcasting.

编辑2

正如用户@KPM的另一个答案所指出的那样,使用is运算符是正确的方法.
2023-01-11 19:54 回答

_______王晓
您可以将检查和强制转换合并为一个语句:
```
let touch = object.anyObject() as UITouch
if let picker = touch.view as? UIPickerView {
    ...
}
```
然后你可以picker在if块内使用.
2023-01-11 19:55 回答

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