我有点卡住了，我试图将系统给定的线程名称传递给我的函数，这样我就可以打印在函数中工作的当前线程的开始和结束时间，我使用全局变量名称。用户必须在给定的时间间隔内输入一个数字。当我输入1001时，线程名称运行良好，但如果我输入1200或10001之类的数字，则名称将不再适用。我举了输出的例子，输出1不是我想要的，输出2是我需要的。我不确定是什么原因导致了这个名字的改变。如果需要任何额外的信息，我很乐意提供
import os
from posixpath import abspath
import time
import sys
import signal
import threading
import platform
import subprocess
from pathlib import Path
import math
lokot = threading.Lock()
lista = []
name = 0
name3 = 0
def divisor(start,end):
lokot.acquire()
start = time.time()
print('{} started working at {}'.format(name, start))
for i in range(int(start),int(end)+1):
if int(end) % i == 0:
lista.append(i)
end = time.time()
print('{} ended working at {}'.format(name, end))
lokot.release()
def new_lista():
lokot.acquire()
start = time.time()
nlista = []
for i in lista:
if i % 2 == 0:
nlista.append(i)
print(nlista)
print('{} was executed in time frame {}'.format(name3,time.time()-start))
lokot.release()
def f4():
while (True):
print ('Input a non negative number in given range <1000,200000>')
number = input()
if number.isalpha() or not number or int(number) not in range(1000,200000) :
print('Number entered is not in the interval <1000,200000>')
continue
else:
global name
global name3
x = int(number) / 2
t1 = threading.Thread(target=divisor, args=(1, x))
t2 = threading.Thread(target=divisor, args=(1, number))
t3 = threading.Thread(target=nova_lista)
name = t1.name
t1.start()
name = t2.name
t2.start()
name3 = t3.name
t3.start()
t1.join()
t2.join()
t3.join()
break
Input 1:
100001
Output 1:
Thread-1 started working at 1624538800.4813018
Thread-2 ended working at 1624538800.4887686
Thread-2 started working at 1624538800.4892647
Thread-2 ended working at 1624538800.5076165
[2, 4, 8, 10, 16, 20, 40, 50, 80, 100, 200, 250, 400, 500, 1000, 1250, 2000, 2500, 5000, 6250, 10000, 12500, 25000, 50000]
Thread-3 dwas executed in time frame 0.0
Input 2:
1001
Output 2:
Thread-1 started working at 1624538882.90607
Thread-1 ended working at 1624538882.9070616
Thread-2 started working at 1624538882.9074266
Thread-2 ended working at 1624538882.9089162
[2, 4, 8, 10, 20, 40, 50, 100, 200, 250, 500, 1000, 1250, 2500, 5000]
Thread-3 dretva se izvodila 0.0
这不一定有效：
name = t1.name
t1.start()
name = t2.name
在t1线程访问name变量之前，没有什么可以阻止第二次赋值。
Q： 为什么不在创建线程时只分配名称，而不让Threading库分配名称呢？例如。；
t1 = threading.Thread(target=divisor, args=(1, x), name="t1")