start.py代码如下。

import threading
class myThread(threading.Thread):
        def __init__(self, threadID, name):
                threading.Thread.__init__(self)
                self.threadID = threadID
                self.name = name

        def run(self):
                currentThreadname = threading.currentThread()
                print "running in ", currentThreadname

thread = myThread(1,"mythrd")
thread.start()

用python启动两次。

python start.py
running in  <myThread(mythrd, started 140461133485824)>
python start.py
running in  <myThread(mythrd, started 140122860668672)>

run.py代码如下。

import threading
class myThread(threading.Thread):
        def __init__(self, threadID, name):
                threading.Thread.__init__(self)
                self.threadID = threadID
                self.name = name

        def run(self):
                currentThreadname = threading.currentThread()
                print "running in ", currentThreadname

thread = myThread(1,"mythrd")
thread.run()

run.py和start.py只有一行不同。
现在启动 run.py 两次。

python  run.py
running in  <_MainThread(MainThread, started 139854546364160)>
python  run.py
running in  <_MainThread(MainThread, started 139854546364160)>

startandrun.py代码如下。

class myThread(threading.Thread):
    def __init__(self, threadID, name):
        threading.Thread.__init__(self)
        self.threadID = threadID
        self.name = name

    def run(self):
        currentThreadname = threading.currentThread()
        print "running in ", currentThreadname

thread = myThread(1,"mythrd")
thread.start()
thread.run()

现在也启动 startandrun.py 两次。

python  startandrun.py
running in  <myThread(mythrd, started 140317119899392)>
running in  <_MainThread(MainThread, started 140317144454912)>
python  startandrun.py
running in running in  <_MainThread(MainThread, started 139980210505472)>
 <myThread(mythrd, started 139980185949952)>

正如 JohanL 所说:
当运行两个单独的线程时，关于哪个将首先执行的所有赌注都将落空。
您基本上是将调度留给操作系统。 第一次执行startandrun.py，thread.start()先于thread.run()执行，结果输出:

running in  <myThread(mythrd, started 140317119899392)>
running in  <_MainThread(MainThread, started 140317144454912)>

第二次执行startandrun.py，thread.start()是在thread.run()之后执行的，为什么输出结果不是:

running in  <_MainThread(MainThread, started 140317144454912)>
running in  <myThread(mythrd, started 140317119899392)>

代替

running in running in  <_MainThread(MainThread, started 139980210505472)>
 <myThread(mythrd, started 139980185949952)>

最佳答案

发生这种情况是因为您打印值的方式:

print "running in ", currentThreadname

添加逗号类似于:

print 'running in ' # without new line at the end
print currentThreadname

由于这两个函数是同时运行的，所以这里是命令的执行方式:

print 'running in ' # without new line FUNCTION #1
print 'running in ' # without new line FUNCTION #2
print currentThreadName # with new line at the end FUNCTION #1
print currentThreadName # with new line at the end FUNCTION #2

尝试使用一个不带逗号的打印语句来理解它应该如何:

def run(self):
    currentThreadname = threading.currentThread()
    print "running in {}".format(currentThreadname)

这将正常运行，但由于这两个函数同时打印，您可能会得到以下输出:

running in <myThread(mythrd, started 10716)>running in <_MainThread(MainThread, started 12132)>

因此，为了证明这可行，您可以使用 time.sleep() 在两次调用之间使用延迟:

import threading
import time

class myThread(threading.Thread):
    def __init__(self, threadID, name):
        threading.Thread.__init__(self)
        self.threadID = threadID
        self.name = name

    def run(self):
        currentThreadname = threading.currentThread()
        print "running in {}".format(currentThreadname)

thread = myThread(1,"mythrd")
thread.start()
time.sleep(0.1)
thread.run()

现在你可以看到你得到了你想要的输出，因为每个函数打印一次，调用之间有 0.1 秒的延迟:

running in <myThread(mythrd, started 5600)>
running in <_MainThread(MainThread, started 7716)>

编辑:

你的问题正是为什么你应该使用多线程而不是运行同一个线程两次。当您使用多线程时您可以使用thread.join() 等待线程结束然后继续代码，或者您可以使用线程。 lock() 这样你就可以继续你的代码，但锁定一个函数一次由一个线程使用。以下是一些示例:

thread.join():

thread = myThread(1, "mythrd")
thread2 = myThread(2, "thrd2")
thread.start()
thread.join() # code will stop here and wait for thread to finish then continue
thread2.run()

threading.lock():

....
    def run(self):
        with lock: # if one thread uses this lock the other threads have to wait
            currentThreadname = threading.currentThread()
            print "running in ", currentThreadname

thread = myThread(1, "mythrd")
thread2 = myThread(2, "thrd2")
lock = threading.Lock()
thread.start() 
thread2.run()
# code keeps running even if there are threads waiting for the lock

关于python - 线程模块中相同的程序不同的输出，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/45911683/