Ios 快速准备阶段不工作

CalViewController：

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

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
 }
}

图形视图控制器：

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

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
 }
}

我只是不明白为什么以及如何解决这个问题 任何想法都将被接受

UIStoryboardSegue在其目标属性中包含您要转换到的实际视图控制器。您不是将数据传递给该视图控制器，而是将其传递给在PrepareForegue one中创建的全新视图控制器，当您离开函数时，该视图控制器会立即被删除

你可能想要一些大致如下的东西：

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

---

我在没有编译器的情况下输入了它，因此您可能需要对它进行调整。

您能否详细介绍一下：什么是不起作用的，什么是应该起作用的。“准备好了吗？”鳞翅目动物已经解决了~Scott的答案~