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runtimeerror:最大递归深度超过python

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当我尝试运行600个测试的智能卡测试工具的主机程序时,我不断收到此错误“RuntimeError:调用Python对象时超出了最大递归深度”,我在第300次测试后得到此错误,我尝试了“sys.setrecursionlimit( 10000)“并解决了这个问题,但我知道这不是解决这个错误的最佳方法,我怎么能改变我的代码所以我不会遇到这个错误:

def SndRcv(self,request):
    print ">> ", request
    device_api.send(request)
    resp = device_api.receive()
    print "<< ", resp
    self.processResponse(resp)

def processResponse(self, K400Message):
    global mWaitingCardRemoval
    ciMsg = card_interface_response
    ciMsgType = card_interface_response.ci_msg

    if ciMsgType is None:
        print 'weird, malformed protobuf response'
        return
    whichMsg = ciMsgType.WhichOneof('msg')
    print 'msg = ' + str(whichMsg)
    if whichMsg is 'collision':
        self.StartSession()
    elif whichMsg is 'card_removed':
        if ciMsgType.issuer== ci.CARD_INTERFACE_MASK_CxLESS:                
            mWaitingCardRemoval &= ~(ciMsgType.issuer)
            if EndofSession is False:
                self.parseMessage()
            if mWaitingCardRemoval !=0:
                self.parseMessage()
            self.StartSession()
    elif whichMsg is 'waiting_removal':
        if EndofSession is False:
            self.parseMessage()
        else:
            mWaitingCardRemoval |= ciMsgType.issuer
    elif whichMsg is 'card_detected':
        mode = ciMsgType.issuer
        reqMsg = pm.get_Deactivate((ci.CARD_INTERFACE_MASK_ANY)& ~(ciMsgType.issuer))
        self.SendOnly(reqMsg)
        acceptMsg = pm.get_Activate(mode)
        self.SndRcv(acceptMsg)
    elif whichMsg is 'card_ready':
        self.StartLoop(ciMsgType.issuer)
    elif whichMsg is 'rapdu':
        self.processCardAPDUResponse(ciMsgType.issuer, ciMsg.data.encode('hex'))
    elif whichMsg is 'card_not_responding':
        if ciMsgType.issuer == ci.CARD_INTERFACE_MASK_CONTACT:
            self.EndCardSession(ciMsgType.issuer,True)
        else:
            self.EndCardSession(ciMsgType.issuer, False)
    elif whichMsg is 'resp_special':
        if ciMsg.data.encode('hex') > 0:
            logging.info(ciMsg.data.encode('hex'))
        else:
            logging.info("")

1 回答

  • 0

    您使用递归来编写一个固有的迭代过程 . 你_12404487_踩过一系列输入 . 处理完输入并报告响应后,您就完成了它 . 在调用堆栈上有's no reason to keep it'的上下文 . 当您点击最终测试并通过您的千多次调用返回时,您不会对结果或函数状态执行任何操作以返回到主程序 .

    将其重写为简单的迭代 . 你是怎么开始的?你如何从一个测试进展到另一个测试?你怎么知道什么时候你的最外面的循环很可能依赖于一个简单的 while

    # Get first response
    while ciMsgType is not None:
        # Process this response
        # Get next response
    

    那会让你感动吗?

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