ios - 快速准备 segue 不工作

Question

CalViewController:

let brain =CalculatorBrain()
override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    let graph = GraphController()
    graph.brain.record = brain.record
}

图形 View Controller :

var brain = CalculatorBrain() {
    didSet {
        draw.variables = ["M": 1]
        let value = brain.evaluate(using: draw.variables)
        print(value)
//at that time outlet didn't set it prints the value work just fine
    }
}

@IBOutlet weak var graphView: GraphView! {
    didSet {
        draw.variables = ["M": 1]
        let value = brain.evaluate(using: draw.variables)
        print(value)
//when outlet got set it not working it print nothing
 }
}

我只是想不通为什么以及如何解决这个问题 任何想法将不胜感激

Answer 1

UIStoryboardSegue 在其destination 属性 中包含您要转换到的实际 View Controller 。您不是将数据传递给该 View Controller ，而是将其传递给您在 prepareForSegue 中创建的全新 View Controller (当您离开该函数时，它会立即被删除)。

您可能想要以下内容:

override func prepare(for segue: UIStoryboardSegue, sender: Any?) {
    if let graph = segue.destination as? GraphController {
        graph.brain.record = brain.record
    }
}

我是在没有编译器的情况下输入的，因此您可能需要对其进行调整。