Question

我想将self作为instancetype传递给此函数的回调闭包：

extension UIView {
    public func onTap(_ handler: @escaping (_ gesture: UITapGestureRecognizer, _ view: Self) -> Void) -> UITapGestureRecognizer {
        ...
    }
}

let view = UIView.init()
view.onTap { tap, v in
    ...
}

但我收到了一个错误：

Self' is only available in a protocol or as the result of a method in a class; did you mean 'UIView'?

我该怎么做？

Answer 1

这可以非常有效地在Swift中使用协议和扩展，这是完美的方案（按书）：

protocol Tappable { }

extension Tappable { // or alternatively: extension Tappable where Self: UIView {
    func onTap(_ handler: @escaping (UITapGestureRecognizer, Self) -> Void) -> UITapGestureRecognizer {
        return UITapGestureRecognizer() // as default to make this snippet sane
    }
}

extension UIView: Tappable { }

然后为例如：

let button = UIButton.init()
button.onTap { tap, v in
    // v is UIButton...
}

而对于例如：

let label = UILabel.init()
label.onTap { tap, v in
    // v is UILabel...
}

等...

注： 您可以在Apple的Extensions中详细了解Protocols或Swift Programming Language Book。< / em>的

Swift：如何在回调闭包中使用instanceType

1 个答案: