ios - 无法将类型 '() -> Void' 的值分配给类型 '(() -> Void)!'

Question

更新到 xcode 8 beta 6 后，出现错误

Cannot assign value of type '() -> Void' to type '(() -> Void)!'

在下面的函数 block 中，第三行:

    // Button sub-class
public class SCLButton: UIButton {
    var actionType = SCLActionType.none
    var target:AnyObject!
    var selector:Selector!
    var action:(()->Void)!

    public init() {
        super.init(frame: CGRect.zero)
    }

    required public init?(coder aDecoder: NSCoder) {
        super.init(coder:aDecoder)
    }

    override public init(frame:CGRect) {
        super.init(frame:frame)
    }

//    required public init?(coder aDecoder: NSCoder) {
//        fatalError("init(coder:) has not been implemented")
//    }
}
public func addButton(_ title:String, action:()->Void)->SCLButton {
    let btn = addButton(title)
    btn.actionType = SCLActionType.closure
    btn.action = action // here is where the error occurs
    btn.addTarget(self, action:#selector(SCLAlertView.buttonTapped(_:)), for:.touchUpInside)
    btn.addTarget(self, action:#selector(SCLAlertView.buttonTapDown(_:)), for:[.touchDown, .touchDragEnter])
    btn.addTarget(self, action:#selector(SCLAlertView.buttonRelease(_:)), for:[.touchUpInside, .touchUpOutside, .touchCancel, .touchDragOutside] )
    return btn
}

有任何修复建议吗？

Answer 1

您的问题似乎与此有关: SE-0103

尝试将 addButton(_:action:) 的方法头更改为:

public func addButton(_ title:String, action:@escaping ()->Void)->SCLButton {

来自新测试版的诊断消息像往常一样令人困惑和不足，但是让您的属性简单地成为非可选 var action:()->Void = {} 会给您带来更多有用的信息信息。