未反映manager.dict中的值更新

问题描述

我希望以下代码可以打印[{0: 100}]，因为它会执行updateList中的加号一百次反过来，它打印[{0: 0}]有什么问题以及如何解决?

I expect the following code to print [{0: 100}],since it do the plus in updateList for a hundred timeIn turn, it prints [{0: 0}]what's the problem and how to rectify it?

from multiprocessing import Process, Lock, Value,Manager
class myWorker:
    def __init__(self, lock, driver, i):
        self.idx=i
        self.driver=driver
        self.lock=lock

    def run(self):
        self.driver.updateList(self.lock,self.idx)

class driver(object):

    def __init__(self):
        manager=Manager()
        self.lock=Lock()
        self.lst=manager.list()
        self.dict1=manager.dict({0:0})
        self.lst.append(self.dict1)

    def workerrun(self,lock, i):
        worker1=myWorker(lock,self,i)
        worker1.run()

    def run(self):
        D=[Process(target=self.workerrun,args=(self.lock,i)) for i in range(10)]
        for d in D:
            d.start()
        for d in D:
            d.join()

    def updateList(self,l,i):
        with self.lock: # acquire lock
            for j in range(10):
                self.lst[0][0]+=1

            print ("update from", i)

if __name__=='__main__':
    dr=driver()
    dr.run()
    print(dr.lst)

推荐答案

 def updateList(self,l,i):
        with self.lock: # acquire lock
            for j in range(10):
                d = self.lst[0]
                d[0] += 1
                self.lst[0]=d
                print ("update from", i,self.lst[0])

来自文档

注意对dict和list代理中的可变值或项的修改将不会通过管理器传播，因为代理无法知道何时修改其值或项.要修改此类项目，您可以将修改后的对象重新分配给容器代理:

如果您已经拥有dict1，则可以直接进行更新:

Seeing as you already have dict1, you can update directly:

   def updateList(self,l,i):
        with self.lock: # acquire lock
            for j in range(10):
                self.dict1[0]+=1
                print ("update from", i,self.lst[0])

这篇关于未反映manager.dict中的值更新的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！