进程退出了，listen的端口没有释放

场景：进程A创建进程B，进程A listen端口P。进程A退出，进程B仍在，此时进程A listen的端口P没有被释放。

原因：进程创建时，默认是共享资源的。这种情况下，进程A的的端口不会回收，因为文件描述符的引用计数仍在。

解决方案：创建socket时，设置属性，端口资源不可被继承。

zmq代码参考

zmq::fd_t zmq::open_socket (int domain_, int type_, int protocol_)
{
    //  Setting this option result in sane behaviour when exec() functions
    //  are used. Old sockets are closed and don't block TCP ports etc.
#if defined ZMQ_HAVE_SOCK_CLOEXEC
    type_ |= SOCK_CLOEXEC;
#endif

    fd_t s = socket (domain_, type_, protocol_);
#ifdef ZMQ_HAVE_WINDOWS
    if (s == INVALID_SOCKET)
        return INVALID_SOCKET;
#else
    if (s == -1)
        return -1;
#endif

    //  If there's no SOCK_CLOEXEC, let's try the second best option. Note that
    //  race condition can cause socket not to be closed (if fork happens
    //  between socket creation and this point).
#if !defined ZMQ_HAVE_SOCK_CLOEXEC && defined FD_CLOEXEC
    int rc = fcntl (s, F_SETFD, FD_CLOEXEC);
    errno_assert (rc != -1);
#endif

    //  On Windows, preventing sockets to be inherited by child processes.
#if defined ZMQ_HAVE_WINDOWS && defined HANDLE_FLAG_INHERIT
    BOOL brc = SetHandleInformation ((HANDLE) s, HANDLE_FLAG_INHERIT, 0);
    win_assert (brc);
#endif

    return s;
}