Question

我正在学习Python 3中的线程。

这是一个简单的概念示例。我希望每个线程在打印之前等待5秒，因此对线程的调用只有1秒大，它必须返回主线程。

但我的代码输出并不是我所期望的。

我的代码：

import threading, time

class MyThread(threading.Thread):  
    def __init__(self, num):
        threading.Thread.__init__(self)
        self.num = num

    def run(self): 
        time.sleep(5)
        print ("I'm thread number: ", self.num)


print ("I'm the principal thread.")

for i in range(0, 10):
    t = MyThread(i)  
    t.start()      
    t.join(1)     
    print("I'm the principal thread, too.")

输出：

I'm the principal thread.
I'm the principal thread, too.
I'm the principal thread, too.
I'm the principal thread, too.
I'm the principal thread, too.
I'm thread number:  0
I'm the principal thread, too.
I'm thread number:  1
I'm the principal thread, too.
I'm thread number:  2
I'm the principal thread, too.
I'm thread number:  3
I'm the principal thread, too.
I'm thread number:  4
I'm the principal thread, too.
I'm thread number:  5
I'm the principal thread, too.
I'm thread number:  6
I'm thread number:  7
I'm thread number:  8
I'm thread number:  9

预期的输出是这样的：

I'm the principal thread.
I'm the principal thread, too.
I'm the principal thread, too.
I'm the principal thread, too.
I'm the principal thread, too.
I'm thread number:  0
I'm the principal thread, too.
I'm the principal thread, too.
I'm the principal thread, too.
I'm the principal thread, too.
I'm thread number:  1
I'm the principal thread, too.
I'm the principal thread, too.
I'm the principal thread, too.
I'm the principal thread, too.
I'm thread number:  2
I'm the principal thread, too.
...

我做错了什么？

Answer 1

听起来你希望主线程每秒继续打印一次，直到另一个线程完成。但这不会发生，因为在尝试join一次后，主线程将放弃并移动到下一个线程。你应该保持join直到线程结束。

import threading, time

class MyThread(threading.Thread):  
    def __init__(self, num):
        threading.Thread.__init__(self)
        self.num = num

    def run(self): 
        time.sleep(5)
        print ("I'm thread number: ", self.num)


print ("I'm the principal thread.")

for i in range(0, 10):
    t = MyThread(i)  
    t.start()      
    while True:
        t.join(1)
        print("I'm the principal thread, too.")
        if not t.isAlive(): break

Answer 2

根据您的输出，我认为以下逻辑必须有效


import threading, time

class MyThread(threading.Thread):  
    def __init__(self, num):
        threading.Thread.__init__(self)
        self.num = num

    def run(self): 
        time.sleep(5)
        print ("I'm thread number: ", self.num)


print ("I'm the principal thread.")

for i in range(0, 10):
    t = MyThread(i)  
    t.start()      
    t.join(1)
    ## You join Times out here, since the thread didnt finish in 1 sec 
    ##(its sleeping for 5 secs)
    while t.isAlive():
        time.sleep(1)
        print("I'm the principal thread, too.")

Answer 3

for循环中有一秒的时滞。因此，在生成的线程等待5秒之前，“我也是主线程。”被打印4次。到那时，num = 0的线程已经完成等待5秒。所以它打印出来。

for i in range(0, 10):
    t = MyThread(i)  
    t.start()      
    t.join(1)     
    print("I'm the principal thread, too.")

在此块中，请注意t.join(1)。它等待1秒。线程't'完成，但因为线程实际上等待5秒。在开始之后，它继续循环。因此，

I'm thread number:  0
I'm the principal thread, too.
I'm thread number:  1

这指向time.sleep(1)。线程编号1在1秒后产生。产生了第0个线程，等等。

循环只运行10次，因此“我也是主线程”。只打印10次。

I'm the principal thread.
I'm the principal thread, too.
I'm the principal thread, too.
I'm the principal thread, too.
I'm the principal thread, too.
I'm thread number:  0
I'm the principal thread, too.
I'm thread number:  1
I'm the principal thread, too.
I'm thread number:  2
I'm the principal thread, too.
I'm thread number:  3
I'm the principal thread, too.
I'm thread number:  4
I'm the principal thread, too.
I'm thread number:  5
I'm the principal thread, too.

此时，循环结束，但剩余的线程在等待5秒后仍在等待打印出来，因此您看到了：

I'm thread number:  6
I'm thread number:  7
I'm thread number:  8
I'm thread number:  9

输出中的差异问题是每个生成的线程等待5秒。你打电话加入它只有一秒钟。因此，没有线程将在分配的1秒内完成。如果你想真正等待生成的线程完成，你应该这样做：

for i in range(0, 10):
    t = MyThread(i)  
    if t.is_alive():
        t.join()    
    print("I'm the principal thread, too.")

多线程不像我期望的那样工作

3 个答案: