C 如何判断线程是否响应了叫醒电话?
我想直接从输入设备读取键盘输入。从这样一个文件中读取需要root权限,我不需要,也不想,在程序的其余部分 我的计划是以root权限启动程序,然后启动工作线程,最后在主线程中删除root权限。然后,主线程可以发送请求,要求工作线程处理下一个键盘输入 我构建了一个最小的示例:C 如何判断线程是否响应了叫醒电话?,c,linux,multithreading,race-condition,worker,C,Linux,Multithreading,Race Condition,Worker,我想直接从输入设备读取键盘输入。从这样一个文件中读取需要root权限,我不需要,也不想,在程序的其余部分 我的计划是以root权限启动程序,然后启动工作线程,最后在主线程中删除root权限。然后,主线程可以发送请求,要求工作线程处理下一个键盘输入 我构建了一个最小的示例: #include <assert.h> #include <stdbool.h> #include <stdio.h> #include <stdlib.h> #include
#include <assert.h>
#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
#include <threads.h>
cnd_t wakeup;
mtx_t mutex;
enum request {
REQUEST_NOTHING,
REQUEST_PING,
REQUEST_TERMINATION
} request;
int daemon(void *arg) {
(void)arg;
int retval;
retval = mtx_lock(&mutex);
assert(retval == thrd_success);
for(;;) {
request = REQUEST_NOTHING;
retval = cnd_wait(&wakeup, &mutex);
assert(retval == thrd_success);
switch(request) {
case REQUEST_NOTHING:
break;
case REQUEST_PING:
puts("pong.");
break;
case REQUEST_TERMINATION:
retval = mtx_unlock(&mutex);
assert(retval == thrd_success);
return 0;
default:
assert(false);
}
}
}
void send(enum request req) {
int retval;
retval = mtx_lock(&mutex);
assert(retval == thrd_success);
request = req;
retval = mtx_unlock(&mutex);
assert(retval == thrd_success);
// TODO race condition: worker thread my not be listening yet
retval = cnd_signal(&wakeup);
assert(retval == thrd_success);
}
int main() {
int retval;
retval = mtx_init(&mutex, mtx_plain);
assert(retval == thrd_success);
retval = cnd_init(&wakeup);
assert(retval == thrd_success);
thrd_t thread;
retval = thrd_create(&thread, daemon, NULL);
assert(retval == thrd_success);
puts("ping.");
send(REQUEST_PING);
// TODO wait for the worker thread to complete
send(REQUEST_TERMINATION);
retval = thrd_join(thread, NULL);
assert(retval == thrd_success);
cnd_destroy(&wakeup);
mtx_destroy(&mutex);
}
#包括
这会导致竞争条件,因为主线程需要在写入请求之前锁定互斥体,而工作线程也需要在等待信号之前锁定互斥体
主线程不会等待工作线程完成。在本例中,这不是问题,因为(1)主线程在发送下一个请求之前必须等待获得锁,并且(2)可以在退出之前简单地加入线程
但在我的实际程序中,我需要等待。显而易见的解决方案是引入另一对cnd\u t
和mtx\u t
,允许工作线程唤醒主线程。但对于这样一个简单的问题来说,这似乎过于复杂了
我找不到太多的资源来演示C11线程库的使用,而且可能都走错了方向。我希望您能提供一些反馈意见,或许能找到上述问题的解决方案。在您的情况下,您不必在流程中永久保留root
权限。
从输入设备读取数据不需要root
。只有在打开时才检查权限
打开设备后,您的进程可以放弃root
权限并继续执行其工作,以非特权用户的身份从打开的文件描述符中读取,因此您不需要单独的进程或线程
尽早放弃扩展特权是实现对特殊资源的访问的标准过程,同时将安全风险降至最低
另见
在其他情况下,您可能需要永久保留特权。在这种情况下,您将需要提升权限的任务与其他任务分离的想法很好,但您必须使用单独的进程而不是线程。免责声明:而@dunes已经解决了我的问题,我了解到权限是在进程的基础上处理的(感谢@solomon),我发现这个问题没有一个解决比赛条件本身的答案是令人愤怒的
事实证明,我只是误解了如何使用这些cnd\t
变量。诀窍是在启动工作线程之前锁定互斥体,并在等待信号时让它解锁互斥体。当给线程发信号时,锁在给线程发信号之前被获得,并且在请求发出之前不会被释放
已经发出信号了,信号熄灭了
此程序不再具有竞赛条件:
#include <assert.h>
#include <threads.h>
mtx_t mx_wakeup;
cnd_t cd_wakeup;
enum request {
REQ_NOTHING,
REQ_TERMINATE
} request;
int daemon(void *arg) {
(void)arg;
for(;;) {
request = REQ_NOTHING;
int retval = cnd_wait(&cd_wakeup, &mx_wakeup);
assert(retval == thrd_success);
if(request == REQ_TERMINATE) {
return 0;
}
}
}
void send(enum request request_) {
int retval;
// The worker thread will unlock the mutex implicitly when
// waiting for a signal, block until that happens.
retval = mtx_lock(&mx_wakeup);
assert(retval == thrd_success);
request = request_;
retval = cnd_signal(&cd_wakeup);
assert(retval == thrd_success);
// The worker thread needs to lock the mutex before waking up,
// this ensures that it doesn't before receiving the signal.
retval = mtx_unlock(&mx_wakeup);
assert(retval == thrd_success);
}
int main() {
int retval;
retval = mtx_init(&mx_wakeup, mtx_plain);
assert(retval == thrd_success);
retval = cnd_init(&cd_wakeup);
assert(retval == thrd_success);
// The mutex will be unlocked by the worker thread when listening.
retval = mtx_lock(&mx_wakeup);
assert(retval == thrd_success);
thrd_t thread;
retval = thrd_create(&thread, daemon, NULL);
assert(retval == thrd_success);
send(REQ_TERMINATE);
retval = thrd_join(thread, NULL);
assert(retval == thrd_success);
cnd_destroy(&cd_wakeup);
mtx_destroy(&mx_wakeup);
}
#包括
#包括
mtx_t mx_唤醒;
cnd_t cd_唤醒;
枚举请求{
什么都不需要,
请求终止
}请求;
int守护进程(void*arg){
(无效)arg;
对于(;;){
请求=不需要任何东西;
int retval=cnd_wait(&cd_唤醒,&mx_唤醒);
断言(retval==thrd_success);
如果(请求==请求终止){
返回0;
}
}
}
无效发送(枚举请求){
内部检索;
//工作线程将在以下情况下隐式解锁互斥锁:
//等待信号,封锁直到发生。
retval=mtx_锁定(&mx_唤醒);
断言(retval==thrd_success);
请求=请求;
retval=cnd_信号(和cd_唤醒);
断言(retval==thrd_success);
//工作线程需要在唤醒前锁定互斥锁,
//这可确保在接收信号之前不会发生任何故障。
retval=mtx_解锁(&mx_唤醒);
断言(retval==thrd_success);
}
int main(){
内部检索;
retval=mtx_init(&mx_唤醒,mtx_普通);
断言(retval==thrd_success);
retval=cnd_init(&cd_唤醒);
断言(retval==thrd_success);
//侦听时,工作线程将解锁互斥锁。
retval=mtx_锁定(&mx_唤醒);
断言(retval==thrd_success);
螺纹;
retval=thrd_create(&thread,daemon,NULL);
断言(retval==thrd_success);
发送(请求终止);
retval=thrd_join(线程,NULL);
断言(retval==thrd_success);
cnd_销毁(和cd_唤醒);
mtx_销毁(和mx_唤醒);
}
等待工作线程现在只是添加另一个条件变量。只有终止代码
必须采用释放互斥锁,否则主线程将永远等待锁
#include <assert.h>
#include <threads.h>
mtx_t mx_wakeup;
cnd_t cd_wakeup, cd_complete;
enum request {
REQ_NOTHING,
REQ_TERMINATE
} request;
int daemon(void *arg) {
(void)arg;
int retval;
for(;;) {
request = REQ_NOTHING;
retval = cnd_wait(&cd_wakeup, &mx_wakeup);
assert(retval == thrd_success);
// The request can be processed here.
// Inform the main thread that the request was completed. The main
// thread can choose to wait or not.
retval = cnd_signal(&cd_complete);
assert(retval == thrd_success);
// Termination is different because the mutex wouldn't be released
// by the next `cnd_wait`, and must happend after the signal was send.
if(request == REQ_TERMINATE) {
retval = mtx_unlock(&mx_wakeup);
assert(retval == thrd_success);
return 0;
}
}
}
void send(enum request request_) {
int retval;
retval = mtx_lock(&mx_wakeup);
assert(retval == thrd_success);
request = request_;
retval = cnd_signal(&cd_wakeup);
assert(retval == thrd_success);
// This unlocks the mutex thus allowing the worker thread to process the
// request, thus the mutex can be reused here.
retval = cnd_wait(&cd_complete, &mx_wakeup);
assert(retval == thrd_success);
retval = mtx_unlock(&mx_wakeup);
assert(retval == thrd_success);
}
int main() {
int retval;
retval = mtx_init(&mx_wakeup, mtx_plain);
assert(retval == thrd_success);
retval = cnd_init(&cd_wakeup);
assert(retval == thrd_success);
// Remember to initialize the new conditional variable.
retval = cnd_init(&cd_complete);
assert(retval == thrd_success);
retval = mtx_lock(&mx_wakeup);
assert(retval == thrd_success);
thrd_t thread;
retval = thrd_create(&thread, daemon, NULL);
assert(retval == thrd_success);
send(REQ_TERMINATE);
retval = thrd_join(thread, NULL);
assert(retval == thrd_success);
cnd_destroy(&cd_wakeup);
mtx_destroy(&mx_wakeup);
}
#包括
#包括
mtx_t mx_唤醒;
cnd_t cd_唤醒,cd_完成;
枚举请求{
什么都不需要,
请求终止
}请求;
int守护进程(void*arg){
(无效)arg;
内部检索;
对于(;;){
请求=不需要任何东西;
retval=cnd_wait(&cd_唤醒,&mx_唤醒);
断言(retval==thrd_success);
//请求可以在这里处理。
//通知主线程请求已完成。主线程
//线程可以选择等待或不等待。
retval=cnd_信号(&cd_完成);
断言(retval==thrd_success);
//终止是不同的,因为互斥锁不会被释放
//在下一个“cnd_wait”之前,必须在信号发送后发生。
如果(请求==请求终止){
retval=mtx_解锁(&mx_唤醒);
断言(retval==thrd_success);
返回0;
}
}
}
无效发送(枚举请求){
内部检索;
retval=mtx_锁定(&mx_唤醒);
断言(retval==thrd_success);
请求=请求;
retval=cnd_信号(和cd_唤醒);
断言(retval==thrd_success);
//这将解锁互斥锁,从而允许工作线程处理
//请求,因此可以在此处重用互斥体。
retval=cnd_wait(&cd_complete,&mx_wakeup);
断言(retval==thrd_su