从多线程函数中获取 return

Question

我正在制作一个 IP 扫描器，它适用于多线程。

在运行宁线程之后，我打印返回值（列表）

但是函数再次运行ning 没有返回列表。

当函数等于一个变量时发生。这需要更多时间，因为函数运行一个接一个。

怎么办？

def first_50IP() :
    
    prefix = "192.168.1."
    condition = "Destination host unreachable"
    condition2 = "Request timed out"
    list1 = []

    for xxx in range(0,50) :  

        ip = prefix + str(xxx)
        code = "ping " + ip + " -n 1 -l 1"    
        ping = os.popen(code).read()
        #ping = subprocess.check_output(code).decode('utf-8')
    
        if condition not in ping and condition2 not in ping:
            print(G + ip)  
            list1.append(ip)
    return list1
                     
def second_50IP() :

    prefix = "192.168.1."
    condition = "Destination host unreachable"
    condition2 = "Request timed out"
    list2 = []

    for xxx in range(50,100) :
      
        ip = prefix + str(xxx)
        code = "ping " + ip + " -n 1 -l 1"    
        ping = os.popen(code).read()
        #ping = subprocess.check_output(code).decode('utf-8')
    
        if condition not in ping and condition2 not in ping:
            print(G + ip)
            list2.append(ip)
    return list2

thread1 = threading.Thread(target=first_50IP) 
thread2 = threading.Thread(target=second_50IP) 

thread1.start() 
thread2.start() 
thread1.join()  
thread2.join()  

print("\nResults : \n")
final_ip1 = first_50IP()
final_ip2 = second_50IP()
print(final_ip1)
print(final_ip2)

Answer 1

你运行同样的功能两次。第一次在专用线程，然后又在主线程。

使用executors and futures来避免：

with concurrent.futures.ThreadPoolExecutor(max_workers=5) as executor:
   first = executor.submit(first_50IP)
   second = executor.submit(second_50IP)

print("\nResults : \n")
final_ip1 = first.result()
final_ip2 = second.result()
print(final_ip1)
print(final_ip2)

旁注：您可以只有一个功能，而不是像这样使用两个几乎相同的功能：

def ping_many(start, count) :
    
    prefix = "192.168.1."
    condition = "Destination host unreachable"
    condition2 = "Request timed out"
    list1 = []

    for xxx in range(start, start+count) :  

        ip = prefix + str(xxx)
        code = "ping " + ip + " -n 1 -l 1"    
        ping = os.popen(code).read()
        #ping = subprocess.check_output(code).decode('utf-8')
    
        if condition not in ping and condition2 not in ping:
            print(G + ip)  
            list1.append(ip)
    return list1

在这种情况下，您可以像这样向执行器提交具有不同参数的函数：

tasks = {}
with concurrent.futures.ThreadPoolExecutor(max_workers=5) as executor:
   for start in [0, 50]:
      tasks.add(executor.submit(ping_many, start, 50))
   
print("\nResults : \n")
for task in tasks:
   print(task.result())

从多线程函数中获取 return

get return from multi-threaded functions

python

python-multithreading

python-3.8