函数内部的函数保留周期

Question

我想知道如何避免在以下情况下保留周期：

private func setupDismissCallbacks() {
// inner func     
func dismiss() {
self.videoExporter?.cancel()
self.rootViewController.dismiss(animated: true, completion: nil)
self.delegate?.childCoordinatorDidFinish(self)
}
// first clousre       
saveModalViewController.onButtonDismiss = {  [weak self] in
// not really using `self` here
guard let self = self else { return }
dismiss()
}
// second clousre  
saveModalViewController.onDimmedAreaDismiss = { [weak self] in
// not really using `self` here
guard let self = self else { return }
dismiss()
}
}

我有一个函数setupDismissCallbacks，它侦听来自saveModalViewControllerself 属性的两个回调。dismiss()是setupDismissCallbacks内部的一个内部函数，我用它来访问self值。

但是在闭包onButtonDismiss和onDimmedAreaDismiss中，我无法访问self来调用dismiss，并且我无法将[weak self]添加到dismiss函数中，因为它是一个函数。

如何验证dismiss中的调用是否不会导致保留周期？

Answer 1

只需将闭包分配给局部变量即可。此时，提取dismiss是没有意义的，因此只需内联它：

private func setupDismissCallbacks() {     
let dismissCallback: () -> Void = {  [weak self] in
guard let self = self else { return }
self.videoExporter?.cancel()
self.rootViewController.dismiss(animated: true, completion: nil)
self.delegate?.childCoordinatorDidFinish(self)
}
saveModalViewController.onButtonDismiss = dismissCallback
saveModalViewController.onDimmedAreaDismiss = dismissCallback
}

Answer 2

>@Alexander评论中解释了这个问题：

您的内部函数捕获self。您的两个闭合捕捉到了这一点 内部函数(包括它所包含的上下文，其中包括self)。由于您的两个闭包被强烈引用saveModalViewController(我认为它被强烈引用self)，您有一个保留周期。

您可以通过不dismiss捕获self来打破此循环。 将SaveCoordinator传递给dismiss：

private func setupDismissCallbacks() {     
func dismiss(_ coordinator: SaveCoordinator) {
coordinator.videoExporter?.cancel()
coordinator(animated: true, completion: nil)
coordinator.delegate?.childCoordinatorDidFinish(coordinator)
}
// first clousre       
saveModalViewController.onButtonDismiss = {  [weak self] in
guard let self = self else { return }
dismiss(self)
}
// second clousre  
saveModalViewController.onDimmedAreaDismiss = { [weak self] in
guard let self = self else { return }
dismiss(self)
}
}