蟒蛇subprocess32与超时，overflowerror

Question

这是我的日志

File "/opt/ibm/db2-governor/helpers/utils.py", line 10, in run_cmd 
output = proc.communicate(timeout = timeout)[0] 
    File "/opt/ibm/dynamite/python/lib/python2.7/site-packages/subprocess32.py", line 927, in communicate 
    stdout, stderr = self._communicate(input, endtime, timeout) 
    File "/opt/ibm/dynamite/python/lib/python2.7/site-packages/subprocess32.py", line 1713, in _communicate 
    orig_timeout) 
    File "/opt/ibm/dynamite/python/lib/python2.7/site-packages/subprocess32.py", line 1786, in _communicate_with_poll 
    ready = poller.poll(self._remaining_time(endtime)) 
OverflowError: Python int too large to convert to C lon 

使触发此代码是

output = proc.communicate(timeout = timeout)[0] 

超时设置为20，出现这种情况intermitently（几乎没有，但它发生） ，即时通讯使用python 2.7.11与subprocess32库，这是一个Python的错误？

好吧，我检查subprocess32.py，行是这样的

endtime = time.time() + timeout 

ready = poller.poll(self._remaining_time(endtime)) 

所以基本上时间戳太大，转换成C INT，有什么我可以做，以解决这个问题？

Answer 1

听起来像一个错误没事。

如果你有兴趣，这里有一个解决方法建议：不是communicate，从工艺stdout在一个线程读取和检查过程结束由要么没有更多的阅读或通过poll返回代码产量。由于你控制循环，你可以在主线程中等待1秒，并为超时倒计时（不需要额外的准确性，因为sleep可以漂移，但这会很简单）。到达0

import threading 

output = "" 

def subp(p): 
    global output 

    while True: 
     # read blocks but since we're in a thread it doesn't matter 
     data = proc.stdout.read() 
     if not data or proc.poll() != None: 
      break 
     output += data 

# here create the process  
proc = subprocess... 

# create a thread, pass the process handle 
t = threading.Thread(target=subp,args=(proc,)) 

while True: 
    if proc.poll() != None: 
     # exit: OK 
     break 
    timeout -= 1 
    if timeout < 0: 
     # took too long: kill 
     proc.terminate() 
     break 
    time.sleep(1) 

t.join()