swift - 将不同类型传递给函数

Question

我正在尝试根据他们想要将其保存到的列表保存到不同的核心数据对象。

是否可以制作一个函数，您传入一个类型并根据该类型将其保存到正确的列表中？

 let context = (UIApplication.shared.delegate as! AppDelegate).persistentContainer.viewContext
    if index == 0 {
        let task = TodayTask(context: context)

        task.name = taskText.text! // Do this in function insead
        task.adress = addressField.text // Do this in function insead
        task.date = userDate // Do this in function insead

    }else if listIndex.selectedSegmentIndex == 1 {
        let task = WeekTask(context: context)
        task.name = taskText.text!
    }else {
        let task = Task(context: context)
        task.name = taskText.text!

func saveToCoreData(task: SomeList){
    let task = SomeList(context: context) // The correct list based on what the user chooses
    task.name = taskText.text!
    task.adress = addressField.text
    task.date = userDate
}

Answer 1

您应该使用一种称为动态分派(dispatch)的技术。例如:

func accept(value: String) {
    //This to process a string
}

func accept(value: Int){
    //This called to process an Int
}

然后你只需调用 accept(value: myValue) 函数就会动态地分配给正确的函数。