Simpy;如何合併未知數量的中斷

Question

我正在使用python和simpy進行仿真。在模擬中，一個實例（中斷）可以被另一箇中斷（中斷者）中斷。我爲每個中斷使用嵌套嘗試除外語句。如果我知道中斷的最大數量，嵌套嘗試除了語句工作。

問題是我不知道會發生多少次中斷（可能是1,2,3，...）。我不知道如何處理一箇中斷未知次數的物體。

下面的代碼適用於三個中斷，但是如果包含第四個中斷（由於三個嵌套嘗試除外語句）會中斷。

是否有可能使代碼更通用，因此它可以處理未知數量的中斷？

任何幫助，非常感謝。

代碼：使用

uninterrupted at 0 
first interrupt 
interrupted at 1 
second interrupt 
interrupted at 2 
third interrupt 
interrupted at 3 
Traceback (most recent call last): 
File "interrupt.py", line 58, in <module> 
    env.run(until=15) 
File "/usr/local/lib/python2.7/dist-packages/simpy/core.py", line 137, in run 
    self.step() 
File "/usr/local/lib/python2.7/dist-packages/simpy/core.py", line 229,  in step 
    raise exc 
simpy.events.Interrupt: Interrupt('fourth interrupt') 

版本：：

import simpy 
import random 

class Interupted(object): 

    def __init__(self, env): 
     self.env = env 
     self.isInterrupted = False 
     self.action = env.process(self.run()) 

    def run(self): 
     self.isInterrupted = False 
     try: 
      print('uninterrupted at %s' % (self.env.now)) 
      yield self.env.timeout(3) 
     except simpy.Interrupt as interrupt: 
      print(interrupt.cause) 
      try: 
       self.isInterrupted = True 
       print('interrupted at %s' % (self.env.now)) 
       yield self.env.timeout(10) 
      except simpy.Interrupt as interrupt: 
       print(interrupt.cause) 
       try: 
        self.isInterrupted = True 
        print('interrupted at %s' % (self.env.now)) 
        yield self.env.timeout(10) 
       except simpy.Interrupt as interrupt: 
        print(interrupt.cause) 
        self.isInterrupted = True 
        print('interrupted at %s' % (self.env.now)) 
        yield self.env.timeout(10) 

class Interruptor(object): 

    def __init__(self, env, interrupted): 
     self.env = env 
     self.interrupted = interrupted 
     self.action = env.process(self.run(interrupted)) 

    def run(self, interrupted): 
     yield self.env.timeout(1) 
     interrupted.action.interrupt("first interrupt") 
     yield self.env.timeout(1) 
     interrupted.action.interrupt("second interrupt") 
     yield self.env.timeout(1) 
     interrupted.action.interrupt("third interrupt") 
     yield self.env.timeout(1) 
     interrupted.action.interrupt("fourth interrupt") 

env = simpy.Environment() 
interrupted = Interupted(env) 
interruptor = Interruptor(env, interrupted) 
env.run(until=15) 

輸出

• 的Python：2.7.3
• 的simpy：3.0.7

Answer 1

我取得了一些進展，並提出了一個解決方案。

多箇中斷不需要嵌套嘗試except語句。單獨的陳述似乎也有效。在發現一些線索和錯誤後，我發現它也可以使用單獨的try語句。

第一個中斷啓動一個計數器。每一箇中斷都會增加計數器，while循環確保所有的中斷都被處理。只要except不包含如上所述的額外的yield語句，就可以工作。

代碼：

import simpy 
import random 

class Interupted(object): 

    def __init__(self, env): 
     self.env = env 
     self.isInterrupted = False 
     self.interruptions = 0 
     self.action = env.process(self.run()) 

    def run(self): 
     print('start at time %s' % (self.env.now)) 
     try: 
      yield self.env.timeout(10) 
     except simpy.Interrupt as interrupt: 
      self.isInterrupted = not self.isInterrupted 
      self.interruptions += 1 
      print('interrupted at time %s interrupted: %s interrupted by: %s' % (self.env.now, self.isInterrupted, interrupt.cause)) 
     while (self.interruptions > 0): 
      self.interruptions = self.interruptions - 1 
      try: 
       yield self.env.timeout(5) 
      except simpy.Interrupt as interrupt: 
       self.interruptions += 1 
       print('interrupted at time %s interrupted: %s interrupted by: %s' % (self.env.now, self.isInterrupted, interrupt.cause)) 
     print('end at time %s' % (self.env.now)) 


class Interruptor(object): 

    def __init__(self, env, interrupted): 
     self.env = env 
     self.interrupted = interrupted 
     self.action = env.process(self.run(interrupted)) 

    def run(self, interrupted): 
     for i in range(16): 
      yield self.env.timeout(5) 
      if(not interrupted.action.processed): 
       interrupted.action.interrupt("interrupt nr: %s" % i) 

env = simpy.Environment() 
interrupted = Interupted(env) 
interruptor = Interruptor(env, interrupted) 
env.run(until=100)