Swift中的可选闭包属性

问题描述

如何在Swift中将可选闭包声明为属性？

How do you declare an optional closure as a property in Swift?

我正在使用此代码：

    var respondToButton:(sender: UIButton) -> Bool

但编译器抱怨该属性未在初始化程序结束时初始化。我相信我可以通过将var声明为可选来解决此问题，但是，我找不到正确的语法。

but the compiler complains that the property is not initialized by the end of the initializer. I believe I can solve this issue by declaring the var as an optional, however, I can not find the correct syntax.

如何将此闭包属性声明为可选项？

How do I declare this closure property as an optional?

推荐答案

我相信你只需要在括号中包含闭包类型，如下所示：

I believe you just need to wrap the closure type in parenthesis, like so:

var respondToButton:((sender: UIButton) -> Bool)?

---

或者，如果这是封闭类型，那么你就是经常使用你可以创建使其更具可读性：

---

Alternatively if this is a closure type you're going to use often you can create a typealias to make it more readable:

typealias buttonResponder = (sender: UIButton) -> Bool

然后在你的班级：

var respondToButton:buttonResponder?

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