Swift在python中等效的可选闭包

Question

我是Python的新手，在Objective-C和Swift方面有很强的背景。

在swift中，您可以创建可用作回调可选闭包。下面是一个示例：

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()

正如我们所看到的，当我们调用'superPower.run()‘时，输出将是The result is: true

在Python中有没有对应的模式？

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()

我用if self.didSuccess is not None and callable(self.didSuccess)仔细检查了这个属性，以确保该属性是可调用的。

Answer 2

这并没有真正的支持。您可以编写如下内容

didSuccess(True) if didSuccess else None

在shell中，这看起来像这样：

>>> didSuccess = None
>>> didSuccess(True) if didSuccess else None
>>> didSuccess = lambda b: print("The result is: {}".format(b))
>>> didSuccess(True) if didSuccess else None
The result is: True