Python：使用短命鍵的線程安全字典，這是正確的嗎？

Question

import threading 
import weakref 
_mainlock = threading.RLock() 
_job_locks = weakref.WeakValueDictionary() 
def do_thing(job_id): 
    _mainlock.acquire() #Dictionary modification lock acquire 
    _job_locks.setdefault(job_id, threading.RLock()) #Possibly modifies the dictionary 
    _mainlock.release() 
    _job_locks[job_id].acquire() 
    try: 
     one_time_init(job_id) 
    finally: 
     _job_locks[job_id].release() 
    #On function return, the weakref.WeakValueDictionary should cause the key to evaporate 

假設do_thing（）在許多線程上被調用很多次，其ID號可能相同也可能不相同（例如，ID 3爲4次，ID爲1次），這個線程是否安全？ one_time_init（）會一次爲特定作業ID運行多次？ （PS：one_time_init保存它的狀態是已經爲每個ID運行一次，因此調用它，如果它已經運行到完成無操作）

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更新後的代碼（感謝THC4k）：

import threading 
import weakref 
_mainlock = threading.RLock() 
_job_locks = weakref.WeakValueDictionary() 

def do_thing(job_id): 
    with _mainlock: 
     jl = _job_locks.setdefault(job_id, threading.RLock()) 
    with jl: 
     one_time_init(job_id) 

Answer 1

好像很安全。爲什麼你甚至需要_job_locks如果one_time_init再次檢查它是否運行？你可以在那裏添加鎖。爲什麼RLock而不是Lock（該函數似乎永遠不會重新輸入）？

不管怎麼說，在with語句如下方式更好：

import threading 
import weakref 
_mainlock = threading.RLock() 
_job_locks = weakref.WeakValueDictionary() 

def do_thing(job_id): 
    with _mainlock: 
     _job_locks.setdefault(job_id, threading.RLock()) 
    with _job_locks[job_id]: 
     one_time_init(job_id)