我正在用fifo编写一个简单的客户端/服务器通信,但我仍然坚持使用信号处理程序来处理客户端请求。
服务器以只读和非阻塞模式打开fifo,读取接收的数据并将一些数据写回客户端fifo。
当服务器端没有信号处理程序时,这实际上工作正常。这是双方的主要代码。
服务器:
int main(int argc, char *argv[])
{
// install handler
struct sigaction action;
action.sa_handler = requestHandler;
sigemptyset(&(action.sa_mask));
action.sa_flags = SA_RESETHAND | SA_RESTART;
sigaction(SIGIO, &action, NULL);
if(!makeFifo(FIFO_READ, 0644))
exit(1);
int rd_fifo = openFifo(FIFO_READ, O_RDONLY | O_NONBLOCK); // non blocking
if(rd_fifo == -1)
exit(1);
// wait for request and answer
while (1) {
qWarning() << "waiting client...";
sleep(1);
QString msg = readFifo(rd_fifo);
qWarning() << "msg = " << msg;
if(msg == "ReqMode") {
int wr_fifo = openFifo(FIFO_WRITE, O_WRONLY); // blocking
writeFifo(wr_fifo, QString("mode"));
break;
} else
qWarning() << "unknow request ..";
}
close(rd_fifo);
unlink(FIFO_READ);
return 0;
}
客户:
int main(int argc, char *argv[])
{
int wr_fifo = openFifo(FIFO_WRITE, O_WRONLY);
if(wr_fifo == -1)
exit(1);
// create a fifo to read server answer
if(!makeFifo(FIFO_READ, 0644))
exit(1);
// ask the server his mode
writeFifo(wr_fifo, QString("ReqMode"));
// read his answer and print it
int rd_fifo = openFifo(FIFO_READ, O_RDONLY); // blocking
qWarning() << "server is in mode : " << readFifo(rd_fifo);
close(rd_fifo);
unlink(FIFO_READ);
return 0;
}
一切都按预期工作(即使没有正确处理所有错误,这只是一个示例代码,以证明这是可能的。)
问题在于,当客户端将数据写入fifo时,处理程序(此处未显示,但只在终端上打印带有接收信号的消息)从不被调用。此外,我已经检查过,如果我从bash(或其他地方)向服务器发送kill -SIGIO
,则执行信号处理程序。
感谢您的帮助。
答案 0 :(得分:1)
实际上,我错过了服务器端的以下3行:
fcntl(rd_fifo, F_SETOWN, getpid()); // set PID of the receiving process
fcntl(rd_fifo, F_SETFL, fcntl(rd_fifo, F_GETFL) | O_ASYNC); // enable asynchronous beahviour
fcntl(rd_fifo, F_SETSIG, SIGIO); // set the signal that is sent when the kernel tell us that there is a read/write on the fifo.
最后一点很重要,因为在我的情况下发送的默认信号是0,所以我必须将它明确地设置为SIGIO以使事情有效。这是服务器端的输出:
waiting client...
nb_read = 0
msg = ""
unknow request ..
waiting client...
signal 29
SIGPOLL
nb_read = 7
msg = "ReqMode"
现在,我想通过将while循环中的内容移动到requestHandler函数中来处理处理程序内的请求是可能的。