我正在尝试使用地图系统来存储和更新聊天服务器的数据。应用程序是多线程的,并使用锁系统来防止多线程访问数据。
问题是:当一个客户端从映射中单独删除时,它是可以的。然而,当我尝试调用多个close时,它会在内存中留下一些。如果我在任何时候调用map上的::clear(),它会导致调试断言错误,或者是"Iterator不兼容“或类似的错误。代码将在第一次运行(使用作为测试连接的80+控制台进行测试),但由于它会留下块,因此不会再次运行。我已经尝试过研究方法,并且我已经编写了系统来停止代码执行,直到每个进程完成。我非常感谢到目前为止的任何帮助,我已经附上了相关的代码片段。
//portion of server code that handles shutting down
DWORD WINAPI runserver(void *params) {
runserverPARAMS *p = (runserverPARAMS*)params;
/*Server stuff*/
serverquit = 0;
//client based cleanup
vector<int> tokill;
map<int,int>::iterator it = clientsockets.begin();
while(it != clientsockets.end()) {
tokill.push_back(it->first);
++it;
}
for(;;) {
for each (int x in tokill) {
clientquit[x] = 1;
while(clientoffline[x] != 1) {
//haulting execution until thread has terminated
}
destoryclient(x);
}
}
//client thread based cleanup complete.
return 0;
}
//clientioprelim
DWORD WINAPI clientioprelim(void* params) {
CLIENTthreadparams *inparams = (CLIENTthreadparams *)params;
/*Socket stuff*/
for(;;) {
/**/
}
else {
if(clientquit[inparams->clientid] == 1)
break;
}
}
clientoffline[inparams->clientid] = 1;
return 0;
}
int LOCKED; //exported as extern via libraries.h so it's visible to other source files
void destoryclient(int clientid) {
for(;;) {
if(LOCKED == 0) {
LOCKED = 1;
shutdown(clientsockets[clientid], 2);
closesocket(clientsockets[clientid]);
if((clientsockets.count(clientid) != 0) && (clientsockets.find(clientid) != clientsockets.end()))
clientsockets.erase(clientsockets.find(clientid));
if((clientname.count(clientid) != 0) && (clientname.find(clientid) != clientname.end()))
clientname.erase(clientname.find(clientid));
if((clientusername.count(clientid) != 0) && (clientusername.find(clientid) != clientusername.end()))
clientusername.erase(clientusername.find(clientid));
if((clientaddr.count(clientid) != 0) && (clientaddr.find(clientid) != clientaddr.end()))
clientaddr.erase(clientusername.find(clientid));
if((clientcontacts.count(clientid) != 0) && (clientcontacts.find(clientid) != clientcontacts.end()))
clientcontacts.erase(clientcontacts.find(clientid));
if((clientquit.count(clientid) != 0) && (clientquit.find(clientid) != clientquit.end()))
clientquit.erase(clientquit.find(clientid));
if((clientthreads.count(clientid) != 0) && (clientthreads.find(clientid) != clientthreads.end()))
clientthreads.erase(clientthreads.find(clientid));
LOCKED = 0;
break;
}
}
return;
}发布于 2012-05-27 10:28:59
您是否真的在使用int进行锁定,或者只是简化了代码?如果你真的使用了一个int:这是行不通的,如果两个线程都在一个线程赋值之前检查变量,那么临界区可以同时输入两次(或更多)(简化)。有关参考,请参阅mutexes in Wikipedia。您可以使用int或boost thread提供的某种互斥锁来代替windows。
https://stackoverflow.com/questions/10771152
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