python - python 中的 Swift 可选闭包等价物

Question

我是 Python 新手，在 Objective-C 和 Swift 方面有深厚的背景。

在 swift 中，您可以创建可用作回调的可选闭包。这是一个例子:

class Process {
  // The closure that will be assigned by the caller of Process.
  var didSuccess: ((Bool)->())?

  func run() {
    let isSuccess = true
    didSuccess?(isSuccess) // If closure is assigned we call it.
  }
}


class Robot {
  private var process = Process()

  init() {
    process.didSuccess = examineProcess // We assign the closure
  }

  func examineProcess(result: Bool) {
    print("The result is: \(result)")
  }

  func run() {
    process.run()
  }
}

let superPower = SuperPower()
superPower.run()

正如我们所见，当我们调用“superPower.run()”时，输出将是 The result is: true

Python 中是否有等效的模式？

Answer 1

Michael Butscher 发布了一个答案，但我对其进行了改进，因为它可能会导致一些错误。

这是我使用的解决方案:

class Process:
  def __init__(self):
    self.didSuccess:  Callable[[bool], None] = None

  def run(self):
    if self.didSuccess is not None and callable(self.didSuccess):
    # we are sure that we will be able to call didSuccess and avoid bugs
    # caused by `myInstance.didSuccess = 3` for example
            self.didSuccess(True)

class Robot:
  def __init__(self):
    self.__process = Process()
    self.__process.didSuccess = examineProcess
    # or lambda
    self.__process.didSuccess = lambda x: print("The result is: ", x)

  func examineProcess(bool, result: bool):
    print("The result is: ", result)


  def run(self):
    self.__process.run()

我使用 if self.didSuccess is not None and callable(self.didSuccess) 仔细检查属性以确保该属性是可调用的。