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我有两个进程,server.c和client.c它们通过POSIX Message Queue进行通信。客户端向队列发送消息,并且mq_notify通知服务器消息已添加到队列中。信号处理程序将接收并处理消息。但是,我无法让它正常工作。从client.c添加一条消息永远不会发送信号处理程序(但是,如果我从server.c添加消息,它会设置处理程序)。服务器仍然可以接收从客户端放入队列中的消息,但由于某些原因,这不会引发server.c的mq_notify中使用的处理程序。任何人都知道这是什么?这里是从各侧的相关示例代码:mq_notify在两个进程之间 - C
client.c
/* queue has already been created, this opens it*/
msgq_id = mq_open(MSGQOBJ_NAME, O_RDWR);
if (msgq_id == (mqd_t)-1) {
perror("In mq_open()");
exit(1);
}
/* sending the message -- mq_send() */
mq_send(msgq_id, packet.mesg_data, strlen(packet.mesg_data), msgprio);
/* closing the queue -- mq_close() */
mq_close(msgq_id);
server.c
void handler()
{
/*for now it just prints that the signal was recieved*/
}
/*i opening the queue -- mq_open() */
msgq_id = mq_open(MSGQOBJ_NAME, O_RDWR);
if (msgq_id == (mqd_t)-1) {
perror("In mq_open()");
exit(1);
}
int main(){
.
.
.
.
/*Set up to be notifed when the queue gets something in it*/
signal(SIGUSR1, handler);
sigevent.sigev_signo = SIGUSR1;;
if(mq_notify (msgq_id, &sigevent) == -1)
{
if(errno == EBUSY)
printf("Another process has registered for notifications.\n");
_exit (EXIT_FAILURE);
}
//strcpy(packet2.mesg_data, "Hello world!");
//mq_send(msgq_id, packet2.mesg_data, strlen(packet2.mesg_data), 0);
while(1)
{
/*wait to be notified*/
}
.
.
.
}
这是否有些事情要和他们在一起单独的进程?
您是否遇到任何错误?程序是否冻结? – 2013-02-28 08:12:40