使用Python os.execl *替换自身进程

Question

I want to replace the self process (running web2py) with different parameter such as bind ip address using Python os.execl* function. Although it success to replace the instance that I found actually it did not KILL the original process but only create a new child process. For example, the original web2py process is running on 10.1.1.1:8000. Then in page's controller it runs,

os.execl('python', 'python', 'web2py.py', '--ip=10.1.1.2', '--port=8000', '--password=')

Ok now we can open the webpage on 10.1.1.2:8000. But if we change back to 10.1.1.1:8000 in the same process again. The new process complain the port has been used by another process.

Is there another way to "REFRESH" web2py service by himself?

Answer 1

os.execl is replacing the current process, but it's retaining the open port; the error message that "the port is in use by another process" is incorrect, it's actually in use by the current process.

On Unix, open file descriptors are inherited across a call to exec , unless the fd is marked close-on-exec with FD_CLOEXEC . This is noted as an issue at http://bugs.python.org/issue12107 .

You should tell web2py to close the port before re-execing your process, or mark the fd as close-on-exec using fcntl module:

fcntl.fcntl(socket, fcntl.F_SETFD, fcntl.fcntl(socket, fcntl.F_GETFD) | fcntl.FD_CLOEXEC)