Swift 4+ swizzle viewWillAppear()

Question

我的问题不是如何swizzle，而是在这个特定的代码片段中发生了什么：

private let swizzling: (UIViewController.Type) -> () = { viewController in

    let originalSelector = #selector(viewController.viewWillAppear(_:))
    let swizzledSelector = #selector(viewController.proj_viewWillAppear(animated:))

    let originalMethod = class_getInstanceMethod(viewController, originalSelector)
    let swizzledMethod = class_getInstanceMethod(viewController, swizzledSelector)

    method_exchangeImplementations(originalMethod, swizzledMethod) }

extension UIViewController {

    open override class func initialize() {
        // make sure this isn't a subclass
        guard self === UIViewController.self else { return }
        swizzling(self)
    }

    // MARK: - Method Swizzling

    func proj_viewWillAppear(animated: Bool) {
        self.proj_viewWillAppear(animated: animated)

        let viewControllerName = NSStringFromClass(type(of: self))
        print("viewWillAppear: \(viewControllerName)")
    } 
 }

代码狙击手来自这里：Swizzling CocoaTouch class

我的问题是关于下面这行代码：

// make sure this isn't a subclass
guard self === UIViewController.self else { return }

为什么我们需要检查它是否不是UIViewController的子类？我的场景是，我想要将带有视图名称的分析数据发送到on(在ever viewWillAppear上)。如果我执行检查，swizzling永远不会起作用，但是当我注释掉这一行时，我会得到我想要的结果，并且每个视图控制器都会发送数据。