python - Python - 全局计数器

Question

我递归地生成几个对象，这些对象需要一个连续的、唯一的 id。如何保证（最简单）python 2.7 中的同步。

iid = 1

def next_id():
    iid += 1
    return iid

def process():
    # .. do something
    id = next_id()

Answer 1

from itertools import count
iid = count()

print next(iid) # 0
print next(iid) # 1
print next(iid) # 2

etc., and

new_iid = count(10)
print next(new_iid) # 10
print next(new_iid) # 11
print next(new_iid) # 12

for starting at other values than 0.

count() is essentially a generator which infinitely yields values.

Answer 2

使用互斥锁：

import threading
iid = 1
iid_lock = threading.Lock()

def next_id():
    global iid
    with iid_lock:
        result = iid
        iid += 1
    return result

您可能希望隐藏类中的内部：

class IdGenerator(object):
    def __init__(self):
        self.cur_id = 1
        self.lock = threading.Lock()
    def next_id(self):
        with self.lock:
            result = self.cur_id
            self.cur_id += 1
        return result

编辑：根据评论，您似乎没有使用线程。这意味着您根本不需要锁定机制。您最初编写的内容就足够了，尽管您需要global关键字来使全局变量可变：

iid = 1
def next_id():
    global iid
    res = iid
    iid += 1
    return res

Answer 3

你在想这样的事情：

 class Counter():
    def __init__(self):
        self.theCount = -1 
    def __call__(self):
        self.theCount += 1
        return self.theCount 

class BorgCounter():
    Borg = {'theCount':-1} 
    def __init__(self):
        self.__dict__ = BorgCounter.Borg 
    def __call__(self):
        self.theCount += 1
        return self.theCount 

myCount = Counter() 
mycount2 = Counter() 
assert(myCount()==0)
assert(mycount2()==0)
assert(mycount2()==1)
assert(myCount()==1)
assert(myCount()==2)

myCount = BorgCounter() 
mycount2 = BorgCounter() 
assert(myCount()==0)
assert(mycount2()==1)
assert(mycount2()==2)
assert(myCount()==3)
assert(myCount()==4)