问题描述
我正在使用 pthread 在我的应用程序上安排一些任务.我从同一个应用程序的旧 C 版本中复制了代码，它运行良好.现在我正在使用 C++ 进行编码，但它不再起作用(基本上它不会触发执行给定函数的 sigevent).即使我使用 timer_gettime，所有创建和启动函数都以 rc 0 退出.
I'm using a pthread to schedule some tasks on my application.I copied the code from my old C version of the same application and it worked perfectly.Now I'm coding with C++ and it doesn't work anymore (basically it doesn't trigger the sigevent executing the given function).All creation and starting functions exit with rc 0, even when i use the timer_gettime.
我简化了很多代码以缩小问题的范围，但我还没有找到:
I simplified a lot the code to narrow down the issue but I could't find it yet:
#include <signal.h>
#include <time.h>
int main (void)
{
char Mytimer[5] = "myt";
timer_t Ti_coarse;
Tcreate(Mytimer, &Ti_coarse, 1000, 1000);
while (1)
{
}
return 0;
static int Tcreate( char *name, timer_t *timerID, int expireMS, int intervalMS )
{
struct sigevent te;
struct itimerspec its;
struct sigaction sa;
int sigNo = SIGRTMIN;
sa.sa_flags = SA_SIGINFO;
sa.sa_sigaction = app;
sigemptyset(&sa.sa_mask);
if (sigaction(sigNo, &sa, NULL) == -1)
{
perror("sigaction");
}
/* Set and enable alarm */
te.sigev_notify = SIGEV_SIGNAL;
te.sigev_signo = sigNo;
te.sigev_value.sival_ptr = timerID;
timer_create(CLOCK_REALTIME, &te, timerID);
its.it_interval.tv_sec = 0;
its.it_interval.tv_nsec = intervalMS * 1000000;
its.it_value.tv_sec = 0;
its.it_value.tv_nsec = expireMS * 1000000;
timer_settime(*timerID, 0, &its, NULL);
return 1;
}
static void app(int sig, siginfo_t *si, void *uc)
{
//do nothing
}
我希望每秒都能看到这是计时器"，但我没有得到任何输出.如果我用 printf("%d", its.it_value) 检查 5 分钟时钟的到期时间，无论过去多久，我总是得到 299.
I expected to see the "THIS IS THE TIMERS" every second, but I don't get any output.If I check the expiration time of a 5 mins clock with printf("%d", its.it_value) I always get 299, no matter how much time passed.
你能帮我找出问题吗?
推荐答案
首先，您没有提供任何人都可以用来重现您的问题的完整示例.请参阅如何创建最小、完整且可验证的示例.
First, you have not provided a full example that anyone can use to reproduce your problem. See How to create a Minimal, Complete, and Verifiable example.
但是您发布的代码确实存在一个严重问题.您不能在信号处理程序中安全地使用 cout.您的信号处理程序可能会导致未定义的行为:
But the code you posted does have one serious problem. You can't safely use cout in a signal handler. Your signal handler can result in undefined behavior:
static void app(int sig, siginfo_t *si, void *uc)
{
cout << "THIS IS THE TIMER" << endl;
}
根据 POSIX 标准:
下表定义了一组函数异步信号安全.因此，应用程序可以调用它们，而无需限制，来自信号捕捉功能....
(在信号处理程序中可以安全调用的大型函数表)
(large table of functions that are safe to call in a signal handler)
实现可能使其他接口异步信号安全.在里面信号的存在，此卷定义的所有功能POSIX.1-2008 在调用或中断时应按照定义的方式运行通过信号捕捉功能，除了当一个信号中断不安全的功能或等效功能(例如处理相当于从初始调用返回后执行的 exit()main()) 和信号捕捉函数调用了一个不安全的函数，行为未定义.
Implementations may make other interfaces async-signal-safe. In the presence of signals, all functions defined by this volume of POSIX.1-2008 shall behave as defined when called from or interrupted by a signal-catching function, with the exception that when a signal interrupts an unsafe function or equivalent (such as the processing equivalent to exit() performed after a return from the initial call to main()) and the signal-catching function calls an unsafe function, the behavior is undefined.
使用 cout 不在异步信号安全函数表中，因此从信号处理程序调用它会导致未定义的行为.
Using cout is not in the table of async-signal-safe functions, so calling it from a signal handler can cause undefined behavior.
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