Python 通过多进程产生的对象标识

首先，当我运行这个(Debian Linux、Python 3.9.7)时，我发现这两个子流程的sss个实例的ID并不相同:

main 140614771273680 <__main__.StateClass object at 0x7fe36d7defd0>
Worker 0 in process 19 (parent process 13): 0, I am 140614770671776: 0
Worker 0 in process 19 (parent process 13): 1, I am 140614770671776: 1
Worker 1 in process 20 (parent process 13): 0, I am 140614761373648: 123
Worker 0 in process 19 (parent process 13): 2, I am 140614770671776: 3
Worker 0 in process 19 (parent process 13): 3, I am 140614770671776: 6
Worker 1 in process 20 (parent process 13): 1, I am 140614761373648: 124
Worker 1 in process 20 (parent process 13): 2, I am 140614761373648: 126
Worker 1 in process 20 (parent process 13): 3, I am 140614761373648: 129
Finish
<__main__.StateClass object at 0x7fe36ce7b7f0> 140614761428976 6
<__main__.StateClass object at 0x7fe36ce7b520> 140614761428256 129
<__main__.StateClass object at 0x7fe36d7defd0> 140614771273680 0

即使您正在Forking 新进程，work_args_list列表中的stato个实例也将作为sss传递给您的工作函数.传递给在不同进程/地址空间中运行的池工作函数的参数由pickle完成，pickle序列化然后反序列化实例，从而生成一个副本，当其反序列化时，in general将具有不同的id.在这种特殊情况下，每个进程在使用fork方法时继承全局变量stato，并且在所有进程/地址空间中都应该具有相同的id.如果我们修改wrapped_work_function以打印出stato的id，我们可以验证这一点，因此:

def wrapped_work_function(a1, a2, sss, qqq):
    print('The id of the inherited stato is', id(stato))
    time.sleep(a1 + 1)
    if a1 == 0:
        sss.state = 0
    else:
        sss.state = 123
    for eee in a2:
        time.sleep(a1 + 1)
        sss.state += eee
        print(
            f"Worker {a1} in process {os.getpid()} (parent process {os.getppid()}): {eee}, {sss()}"
        )
    return sss

则打印输出为:

main 140456701534160 <__main__.StateClass object at 0x7fbe9fcd1fd0>
The id of the inherited stato is 140456701534160
The id of the inherited stato is 140456701534160
Worker 0 in process 43 (parent process 37): 0, I am 140456700920112: 0
Worker 0 in process 43 (parent process 37): 1, I am 140456700920112: 1
Worker 1 in process 44 (parent process 37): 0, I am 140456700920112: 123
Worker 0 in process 43 (parent process 37): 2, I am 140456700920112: 3
Worker 0 in process 43 (parent process 37): 3, I am 140456700920112: 6
Worker 1 in process 44 (parent process 37): 1, I am 140456700920112: 124
Worker 1 in process 44 (parent process 37): 2, I am 140456700920112: 126
Worker 1 in process 44 (parent process 37): 3, I am 140456700920112: 129
Finish
<__main__.StateClass object at 0x7fbe9f36e880> 140456691689600 6
<__main__.StateClass object at 0x7fbe9f36eb20> 140456691690272 129
<__main__.StateClass object at 0x7fbe9fcd1fd0> 140456701534160 0

所有地址空间都会看到stato的相同id，即140456701534160.如果每个地址空间都看到继承stato的相同id，那么sss的id(应该是stato,的单独副本)不能与stato具有相同的id.当我运行代码时，正如我所期望的那样，它们具有不同的ID.但在不同地址空间中运行的每个sss可能彼此具有相同的id，但这并不能保证(在第二次运行时，它们是相同的).

But even if the 100 instances have the same id and the same address, these are two instances existing in two different processes and thus two different address spaces. That is why they can maintain distinct states.另一方面，当您的worker函数返回sss时，它将使用pickle传递回主进程，pickle对实例进行序列化和反序列化，从而实际上生成原始实例的副本.这就是为什么返回的ID不同.

另一方面:您有bullo = result.get(timeout=100)个可能超时的测试.但在这条语句之前，会调用pool.close()和pool.join().这两个调用将等待所有提交的任务完成.因此，当您调用result.get时，任务保证已完成，并且永远不会导致超时异常.