swift - 如何将 Void 类型添加到元组？

Question

我在为元组集合声明 Void 类型时遇到了一些麻烦。我收到以下错误:Nil is not compatible with expected element type 'Void '

var messages: [(person: String, type: String, text: String, action: Void, isSender: Bool)]? = [
    (person: "Buddy", type: "instruction", text: "Wat vervelend van uw schade. De volgende vragen moeten direct op locatie ingevuld worden", action: nil,  isSender: false),
    (person: "User", type: "question", text: "Wat is uw locatie en tijd?", action: askPermissionForGPS(), isSender: true)
]

我已经尝试将 lazy 放在字典集合之前，以及声明类型 Void、Void -> () 的不同方式， () -> Void，但没有结果。

我想要的是元组可以有一个函数作为参数或 nil。

Answer 1

如果你想使用nil，你必须将action参数声明为可选

var messages: [(person: String, type: String, text: String, action: Void?, isSender: Bool)]? = [ ...

但我建议将action参数声明为闭包

var messages: [(person: String, type: String, text: String, action: (()->Void)?, isSender: Bool)]? = [ ...

然后你可以只传递函数名

 (person: "User", 
    type: "question", 
    text: "Wat is uw locatie en tijd?", 
  action: askPermissionForGPS, 
isSender: true)