我有一个很大的问题
我最近开始使用守护进程和套接字编程。我当前的设置包括一个守护进程,它创建一个套接字和一个写入消息并读取已处理答案的客户端
那么,应阅读答案。
守护进程守护自身,创建套接字甚至处理消息(我可以通过我做的日志条目进行验证)但是一旦我尝试将答案发送到客户端,守护进程就会被阻止;客户端永远不会收到答案,如果我ctrl-c
客户端,守护程序也会关闭
这对我来说是完全奇怪的行为,我无法解释发生了什么,这就是为什么我会问你任何提示。
摘要:
- 套接字类型为AF_UNIX
,因此本地且不涉及网络
- 守护程序成功接收消息,但无法写回
- 客户端无法读取(因为服务器的send()
正在阻止)和
如果我ctrl-c
客户端守护进程也退出(但该信号没有被发送到守护进程)
下面是我的代码,被剥离到有趣且无法工作的部分 非常感谢提示,帮助和建议!
更新
还添加了客户端代码!
int main()
{
//those two work just fine
createSocket();
daemonize();
//declare some variables needed in the big loop
FILE *client_fd;
int fd;
socklen_t len;
string tmp;
char *buf = new char[1024];
while(1)
{
struct sockaddr_un saun;
len = sizeof(saun);
//receive an fd for the incoming connection
if((fd = accept(m_socket, (sockaddr*)&saun, &len)) < 0)
{
writeLog("Accepting connection failed!");
continue;
}
//i wanted to use fgets, therefore obtain a FILE object here
client_fd = fdopen(fd, "r");
if(client_fd == NULL)
{
writeLog("fdopen() failed!");
continue;
}
while(fgets(buf, bufferSize, client_fd) != NULL)
{
//message read succeeds (verified)
writeLog(buf);
tmp = evaluateMsg(buf);
//tmp is not empty (verified)
writeLog(tmp);
//send seems to block, nothing happens from now on
//I don't even get in one of the if-else branches
if(send(fd, (tmp + "\n").c_str(), tmp.size() + 1, 0) < 0)
{
writeLog("Could not write message back!");
continue;
}
else
writeLog("bytes written!");
}
close(fd);
}
}
以下是客户端代码:
int main()
{
int sockfd;
sockaddr_un addr;
sockfd = socket(AF_UNIX, SOCK_STREAM, 0);
addr.sun_family = AF_UNIX;
strcpy(addr.sun_path, SOCKET_PATH);
//the connection is successfull
if(connect(sockfd, (sockaddr*)&addr, sizeof(addr)) < 0)
{
cerr<<"Connection failed!"<<endl;
exit(1);
}
//message is written successfully
write(sockfd, "hello", 5);
char *buf = new char[1024];
//same as for the server, I wanted to use fgets
//therefore need to obtain a FILE object
FILE *fd = fdopen(sockfd, "r");
fgets(buf, 1024, fd);
cout<<buf<<endl;
close(sockfd);
}
答案 0 :(得分:0)
好吧,我自己找到了解决方案。
问题是对fgets()
的(方便)调用;我不得不使用read()
,现在一切都像魅力一样
我不确定为什么read()
是强制性的,但我认为它与缓冲区有关,缓冲区只要读取就清空了...
但这只是我的假设!