ios - 如何接受多种类型作为函数中的参数？

Question

我想在另一个 swift 文件中创建一个通用函数，用于使用动画隐藏和显示 UI 对象。这是两个代码:

func hide(object: /*My UI Object*/, duration: Double, delay: Double) {
    UIView.animate(withDuration: duration, delay: delay, options: [], animations: {
        object.alpha = 0
    } ,
                   completion: nil
    )
}

func show(object: /*My UI Object*/, duration: Double, delay: Double) {
    UIView.animate(withDuration: duration, delay: delay, options: [], animations: {
        object.alpha = 1
    } ,
                   completion: nil
    )
}

我想使用以下对象的此功能:UILabel、UIButton、UIView、UITextField 等等。我找不到任何允许多种类型作为“对象”参数的方法。

我还尝试将“object”的类型设置为 Any，但这会导致“object”没有成员“alpha”的错误。

将类型设置为 AnyObject 会导致无法分配给属性的错误，因为“object”是“let”常量。

感谢您的帮助!

Answer 1

您可以将参数键入 UIView，因为所有提到的控件都继承自它。

func hide(object: UIView, duration: Double, delay: Double) {
    UIView.animate(withDuration: duration, delay: delay, options: [], animations: {
        object.alpha = 0
    } ,
                   completion: nil
    )
}

func show(object: UIView, duration: Double, delay: Double) {
    UIView.animate(withDuration: duration, delay: delay, options: [], animations: {
        object.alpha = 1
    } ,
                   completion: nil
    )
}