以下简单脚本会间歇性地挂起subprocess.Popen(大约30%的时间)。
除非use_lock = True,然后它永远不会挂起,导致我相信子进程不是线程安全的!
预期的行为是脚本在5-6秒内完成。
为了演示这个bug,只需运行几次“python bugProof.py”直到它挂起。 Ctrl-C退出。你会看到'后Popen'只出现一次或两次,但不是第三次出现。
import subprocess, threading, fcntl, os, time
end_time = time.time()+5
lock = threading.Lock()
use_lock = False
path_to_factorial = os.path.join(os.path.dirname(os.path.realpath(__file__)),'factorial.sh')
def testFunction():
print threading.current_thread().name, '| pre-Popen'
if use_lock: lock.acquire()
p = subprocess.Popen([path_to_factorial], stdin=subprocess.PIPE, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
if use_lock: lock.release()
print threading.current_thread().name, '| post-Popen'
fcntl.fcntl(p.stdout, fcntl.F_SETFL, os.O_NONBLOCK)
fcntl.fcntl(p.stderr, fcntl.F_SETFL, os.O_NONBLOCK)
while time.time()<end_time:
try: p.stdout.read()
except: pass
try: p.stderr.read()
except: pass
print threading.current_thread().name, '| DONE'
for i in range(3):
threading.Thread(target=testFunction).start()
上面引用的shell脚本(factorial.sh):
#!/bin/sh
echo "Calculating factorial (anything that's somewhat compute intensive, this script takes 3 sec on my machine"
ans=1
counter=0
fact=999
while [ $fact -ne $counter ]
do
counter=`expr $counter + 1`
ans=`expr $ans \* $counter`
done
echo "Factorial calculation done"
read -p "Test input (this part is critical for bug to occur): " buf
echo "$buf"
系统信息:
Linux 2.6.32-358.123.2.openstack.el6.x86_64#1 SMP Thu Sep 26 17:14:58 EDT 2013 x86_64 x86_64 x86_64 GNU / Linux
Python 2.7.3(默认,2013年1月22日,11:34:30)
[bCC 4.4.6 20120305(Red Hat 4.4.6-4)]在linux2上