嵌套通知中的 DispatchGroup [weak self] 处理

Question

我有两个异步函数。我需要运行基于第一个输出的第二个函数和运行第二个完成后的第三个函数。我是这样做的。这是处理序列化的正确方法吗？以及如何处理弱自引用？

let dispatchGroup = DispatchGroup()

dispatchGroup.enter()
function1() // async completion contains leave()

dispatchGroup.notify(queue: DispatchQueue.main) { [weak self] in
    guard let strongSelf = self else {return}
    strongSelf.dispatchGroup.enter()    
    strongSelf.function2() // Also async, dependent on the result of function1. contains leave()

    strongSelf.dispatchGroup.notify(queue: DispatchQueue.main) {
        strongSelf????.function3()
        print("results of function1 and function2")
        // I must wait to finish first two tasks in order, in other words serial queue
    }
}

Answer 1

1）正确方法：在上一个方法完成后调用方法

class SomeAsyncType {

    typealias SomeResultType = String

    func async(_ someDate: Any, completion: @escaping (SomeResultType) -> Void) {
        self.asyncfunctionA(someDate, completion: completion)
    }

    private func asyncfunctionA(_ someDate: Any, completion: @escaping (SomeResultType) -> Void) {
        DispatchQueue.main.async { [weak self] in
            sleep(1)
            print("ResultA")
            self?.asyncfunctionB("ResultA", completion: completion)
        }
    }

    private func asyncfunctionB(_ someDate: Any, completion: @escaping (SomeResultType) -> Void) {
        DispatchQueue.main.async { [weak self] in
            sleep(2)
            print("ResultB")
            self?.asyncfunctionC("ResultB", completion: completion)
        }
    }

    private func asyncfunctionC(_ someDate: Any, completion: @escaping (SomeResultType) -> Void) {
       DispatchQueue.main.async {
            sleep(3)
            print("ResultC")
            completion("")
            print("All completed")
        }
    }
}

用法：

let test = SomeAsyncType() // hold reference 
test.async("noDate") { result in
    print(result)
}

控制台：

ResultA
ResultB
ResultC

All completed

2) 如何处理弱自引用？

这是新范围，您可以再次使用 [weak self]

dispatchGroup.notify(queue: DispatchQueue.main) { [weak self] in
    guard let strongSelf = self else {return}
    strongSelf.dispatchGroup.enter()
    strongSelf.function2() // Also async, dependent on the result of function1. contains leave()

    strongSelf.dispatchGroup.notify(queue: DispatchQueue.main) { [weak self] in
        guard let strongSelf = self else {return}
        strongSelf.function3()
        print("results of function1 and function2")
    }
}

Answer 2

这取决于你的函数 1 和 2 是什么，但我会考虑使用 completionHandlerInstead:

function1(completionHandler: { [weak self] results in
   guard let strongSelf = self else {return} 

   if results == true //if it'd be bool
   {
      strongSelf.function2(completionHandler: { [weak self] results in
   guard let strongSelf = self else {return}

      //here you can get results from both if you want
      })
   }
   else
   {
   //false
   }
})

嵌套通知中的 DispatchGroup [weak self] 处理

DispatchGroup [weak self] handling in nested notify

weak-references

blockingqueue

grand-central-dispatch

swift

swift3

1）正确方法：在上一个方法完成后调用方法

用法：

控制台：

2) 如何处理弱自引用？