I'm building some FastCGI apps and it sort of bugs me that lighttpd doesn't kill them off after they've been idle, so I'm trying to have them close on their own.
I tried using
signal(SIGALRM, close);
alarm(300);
and having the close function execute exit(0), and that works almost well.
The problem is the close function is being called every time the main program loop runs though (I call alarm(300) each loop to reset it). I've read the man page for alarm() and it doesn't seem as though calling it multiple times with the same value should trip SIGALRM so I'm assuming Lighttpd is sending an alarm signal.
The big question! Is there a way to run a method after a specific interval, and have that interval be resettable without SIGALRM? I'd be nice if I could have multiple alarms as well.
Here's the whole app thus far:
#include <stdlib.h>
#include <stdarg.h>
#include <signal.h>
#include "fcgiapp.h"
FCGX_Stream *in, *out, *err;
FCGX_ParamArray envp;
int calls = 0;
void print(char*, ...);
void close();
int main(void)
{
// If I'm not used for five minutes, leave
signal(SIGALRM, close);
int reqCount = 0;
while (FCGX_Accept(&in, &out, &err, &envp) >= 0)
{
print("Content-type: text/plain\r\n\r\n");
int i = 0;
char **elements = envp;
print("Environment:\n");
while (elements[i])
print("\t%s\n", elements[i++]);
print("\n\nDone. Have served %d requests", ++reqCount);
print("\nFor some reason, close was called %d times", calls);
alarm(300);
}
return 0;
}
void print(char *strFormat, ...)
{
va_list args;
va_start(args, strFormat);
FCGX_VFPrintF(out, strFormat, args);
va_end(args);
}
void close()
{
calls++;
// exit(0);
}
the best way is: add a thread so that you can remove signal and alarm, and sync the thread and your main code (main thread).
I'd probably use POSIX timers. Timers do not have to use signals. You have a choice between not notifying at all, raising a signal, or running a function as a new thread (which I would do as it will not interfere with fastcgi).
Make sure you include <signal.h> and <time.h>, and link with -lrt
First, I'd fill out your sigevent structure:
struct sigevent myTimerSignal = {
.sigev_notify = SIGEV_THREAD,
.sigev_notify_function = close //Make sure you change your function declaration to close(union sigval), you do not need to use the sigval unless you store data in your event too
};
Now create your timer:
timer_t myTimer;
if(timer_create(CLOCK_REALTIME, &myTimerSignal, &myTimer)){
//An error occurred, handle it
}
Lets arm it, it will call close() in a new thread in 300 seconds:
struct itimerspec timeUntilClose = {
.it_value = {
.tv_sec = 300 //300 seconds
}
};
if(timer_settime(myTimer, 0, &timeUntilClose, NULL)){
//Handle the error
}
Now, you should have a timer ready to stop the program after 300 seconds. I know I may be late, but I hope this helps a future reader.
Maybe you can wrap the close function by another function which will first call sleep()?
The argument to the alarm call is seconds, not minutes. So you're asking to be woken up in 5 seconds after each time through the main loop.
Here's a solution that sort of avoids the point of the question, but it works. It will respond to only my application's signal events:
void close(int intSignal, siginfo_t *info, void *context)
{
// For some stupid reason MY signal doesn't populate siginfo_t
if (!info)
{
count++;
}
}
If the siginfo struct is empty, that's because alarm() tripped it. If an outside process does it, siginfo_t.si_pid is populated with zero.
I still don't like this solution, but it works. Odd problem now is that doing an exit(0) doesn't close the application, though lighttpd thinks it's gone and spawns another. This means that now I've got rouge processes. raise(SIGUSR1) which is what is supposed to stop FastCGI scripts doesn't seem to do the trick either... hmmm...
Question still remains: How does one call asynchronous functions on an interval timer without the use of signals?
Try to close all the file descriptors (including stdin and stdout). This should close the CGI instance if its idle.
You can use select() with timeout to schedule instead of SIGALRM
Related
So I'm trying to call an alarm to display a message "still working.." every second.
I included signal.h.
Outside of my main I have my function: (I never declare/define s for int s)
void display_message(int s); //Function for alarm set up
void display_message(int s) {
printf("copyit: Still working...\n" );
alarm(1); //for every second
signal(SIGALRM, display_message);
}
Then, in my main
while(1)
{
signal(SIGALRM, display_message);
alarm(1); //Alarm signal every second.
That's in there as soon as the loop begins. But the program never outputs the 'still working...' message. What am I doing incorrectly? Thank you, ver much appreciated.
Signal handlers are not supposed to contain "business logic" or make library calls such as printf. See C11 ยง7.1.4/4 and its footnote:
Thus, a signal handler cannot, in general, call standard library functions.
All the signal handler should do is set a flag to be acted upon by non-interrupt code, and unblock a waiting system call. This program runs correctly and does not risk crashing, even if some I/O or other functionality were added:
#include <signal.h>
#include <stdio.h>
#include <stdbool.h>
#include <unistd.h>
volatile sig_atomic_t print_flag = false;
void handle_alarm( int sig ) {
print_flag = true;
}
int main() {
signal( SIGALRM, handle_alarm ); // Install handler first,
alarm( 1 ); // before scheduling it to be called.
for (;;) {
sleep( 5 ); // Pretend to do something. Could also be read() or select().
if ( print_flag ) {
printf( "Hello\n" );
print_flag = false;
alarm( 1 ); // Reschedule.
}
}
}
Move the calls to signal and alarm to just before your loop. Calling alarm over and over at high speed keeps resetting the alarm to be in one second from that point, so you never reach the end of that second!
For example:
#include <stdio.h>
#include <signal.h>
#include <unistd.h>
void display_message(int s) {
printf("copyit: Still working...\n" );
alarm(1); //for every second
signal(SIGALRM, display_message);
}
int main(void) {
signal(SIGALRM, display_message);
alarm(1);
int n = 0;
while (1) {
++n;
}
return 0;
}
Do not call alarm() twice, just call it once in main() to initiate the callback, then once in display_message().
Try this code on Linux (Debian 7.8) :
#include <stdio.h>
#include <signal.h>
void display_message(int s); //Function for alarm set up
void display_message(int s)
{
printf("copyit: Still working...\n" );
alarm(1); //for every second
signal(SIGALRM, display_message);
}
int main()
{
signal(SIGALRM, display_message);
alarm(1); // Initial timeout setting
while (1)
{
pause();
}
}
The result will be the following one :
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
copyit: Still working...
The alarm() call is for a one off signal.
To repeat an alarm, you have to call alarm() again each time the signal occurs.
Another issue, also, is that you're likely to get EINTR errors. Many system functions get interrupted when you receive a signal. This makes for much more complicated programming since many of the OS functions are affected.
In any event, the correct way to wait for the next SIGALRM is to use the pause() function. Something the others have not mentioned (instead they have tight loops, ugly!)
That being said, what you are trying to do would be much easier with a simple sleep() call as in:
// print a message every second (simplified version)
for(;;)
{
printf("My Message\n");
sleep(1);
}
and such a loop could appear in a separate thread. Then you don't need a Unix signal to implement the feat.
Note: The sleep() function is actually implemented using the same timer as the alarm() and it is clearly mentioned that you should not mix both functions in the same code.
sleep(3) may be implemented using SIGALRM; mixing calls to alarm() and sleep(3) is a bad idea.
(From Linux man alarm)
void alarm_handler(int)
{
alarm(1); // recurring alarm
}
int main(int argc, char *argv[])
{
signal(SIGALRM, alarm_handler);
alarm(1);
for(;;)
{
printf("My Message\n");
// ...do other work here if needed...
pause();
}
// not reached (use Ctrl-C to exit)
return 0;
}
You can create variations. For example, if you want the first message to happen after 1 second instead of immediately, move the pause() before the printf().
The "other work" comment supposes that your other work does not take more than 1 second.
It is possible to get the alarm signal on a specific thread if work is required in parallel, however, this can be complicated if any other timers are required (i.e. you can't easily share the alarm() timer with other functions.)
P.S. as mentioned by others, doing your printf() inside the signal handler is not a good idea at all.
There is another version where the alarm() is reset inside main() and the first message appears after one second and the loop runs for 60 seconds (1 minute):
void alarm_handler(int)
{
}
int main(int argc, char *argv[])
{
signal(SIGALRM, alarm_handler);
for(int seconds(0); seconds < 60; ++seconds)
{
alarm(1);
// ...do other work here if needed...
pause();
printf("My Message\n");
}
// reached after 1 minute
return 0;
}
Note that with this method, the time when the message will be printed is going to be skewed. The time to print your message is added to the clock before you restart the alarm... so it is always going to be a little over 1 second between each call. The other loop is better in that respect but it still is skewed. For a perfect (much better) timer, the poll() function is much better as you can specify when to wake up next. poll() can be used just and only with a timer. My Snap library uses that capability (look for the run() function, near the bottom of the file). In 2019. I moved that one .cpp file to the eventdispatcher library. The run() function is in the communicator.cpp file.
POSIX permits certain of its functions to be called from signal handling context, the async-signal safe functions, search for "async-sgnal safe" here. (These may be understood as "system calls" rather than library calls). Notably, this includes write(2).
So you could do
void
display_message (int s) {
static char const working_message [] = "copyit: Still working...\n";
write (1, working_message, sizeof working_message - sizeof "");
alarm(1); /* for every second */
}
By the way, precise periodic alarms are better implemented using setitimer(2),
since these will not be subject to drift. Retriggering the alarm via software, as done here, will unavoidably accumulate error over time because of the time spent executing the software as well as scheduling latencies.
In POSIX sigaction(2) superceedes signal(2) for good reason:
the original Unix signal handling model was simple. In particular,
a signal handler was reset to its original "deposition" (e.g., terminate
the process) once it was fired. You would have to re-associate
SIGALRM with display_message() by calling signal() just before
calling alarm() in display_message().
An even more important reason for using sigaction(2) is the
SA_RESTART flag. Normally, system calls are interrupted when
a signal handler is invoked. I.e., when then signal handler returns,
the system call returns an error indication (often -1) and errno is
set to EINTR, interrupted system call. (One reason for this
is to be able to use SIGALRM to effect time outs, another is
to have a higher instance, such as a user, to "unblock" the
current process by sending it a signal, e.g.,
SIGINT by pressing control-C at the terminal).
In your case, you want signal handling to be transparent
to the rest of the code, so you would set the SA_RESTART flag
when invoking sigaction(2). This means the kernel should
restart the interrupted system call automatically.
ooga is correct that you keep reloading the alarm so that it will never go off. This works. I just put a sleep in here so you don't keep stepping on yourself in the loop but you might want to substitute something more useful depending on where you are headed with this.
void display_message(int s)
{
printf("copyit: Still working...\n" );
// alarm(1); //for every second
// signal(SIGALRM, display_message);
}
int main(int argc, char *argv[])
{
int ret;
while(1)
{
signal(SIGALRM, display_message);
alarm(1);
if ((ret = sleep(3)) != 0)
{
printf("sleep was interrupted by SIGALRM\n");
}
}
return (0);
}
I am trying to figure out how to get rid of a reliance on the pthread_timedjoin_np because I am trying to build some code on OSX.
Right now I have a Queue of threads that I am popping from, doing that pthread_timedjoin_np and if they dont return, they get pushed back on the queue.
The end of the thread_function that is called for each thread does a pthread_exit(0); so that the recieving thread can check for a return value of zero.
I thought i might try to use pthread_cond_timedwait() to achieve a similar effect, however I think i am missing a step.
I thought I would be able to make worker Thread A signal a condition AND pthread_exit() within a mutex, , and worker Thread B could wake up on the signal, and then pthread_join(). The problem is, Thread B doesn't know which thread threw the conditional signal. Do I need to explicitly pass that as part of the conditonal signal or what?
Thanks
Derek
Here is a portable implementation of pthread_timedjoin_np. It's a bit costly, but it's a full drop-in replacement:
struct args {
int joined;
pthread_t td;
pthread_mutex_t mtx;
pthread_cond_t cond;
void **res;
};
static void *waiter(void *ap)
{
struct args *args = ap;
pthread_join(args->td, args->res);
pthread_mutex_lock(&args->mtx);
args->joined = 1;
pthread_mutex_unlock(&args->mtx);
pthread_cond_signal(&args->cond);
return 0;
}
int pthread_timedjoin_np(pthread_t td, void **res, struct timespec *ts)
{
pthread_t tmp;
int ret;
struct args args = { .td = td, .res = res };
pthread_mutex_init(&args.mtx, 0);
pthread_cond_init(&args.cond, 0);
pthread_mutex_lock(&args.mtx);
ret = pthread_create(&tmp, 0, waiter, &args);
if (!ret)
do ret = pthread_cond_timedwait(&args.cond, &args.mtx, ts);
while (!args.joined && ret != ETIMEDOUT);
pthread_mutex_unlock(&args.mtx);
pthread_cancel(tmp);
pthread_join(tmp, 0);
pthread_cond_destroy(&args.cond);
pthread_mutex_destroy(&args.mtx);
return args.joined ? 0 : ret;
}
There may be small errors since I wrote this on the spot and did not test it, but the concept is sound.
Producer-consumer queue. Have the threads queue *themselves, and so their results,(if any), to the queue before they exit. Wait on the queue.
No polling, no latency.
With your current design, you would have to join() the returned threads get the valueptr and to ensure that they are destroyed.
Maybe you could sometime move to a real threadpool, where task items are queued to threads that never terminate, (so eliminating thread create/terminate/destroy overhead)?
solution with alarm.
pthread should enable cancel, so it can stop by external.(even with pthread_timedjoin_np).
pthread_timedjoin_np return with ETIMEOUT after waited time.
set alarm, use alarm also can give "TIMEOUT" signal.
In handler, just pthread_cancel it. (only timeout run this).
pthread_join it in main thread.
reset alarm
I write test code in here:github
I need help to clear my concepts.
I have a function which toggle the Led status on/off after every second. Now the code for the on/off runs inside infite loop.
Example:
void ToggleLed( int pin_number)
{
// some code
while(1)
{
// code to execute the Led status
}
}
Now when I integrate this code with base line and called that function inside other function it just doesnt work no other functionality of software works.
Question: Function has infinite-loop and that it doesn't come out of control and other functions called after that function doesn't work.
If that is the case do I need to provide separate thread to it?
Any suggestion will be helpful.
Yes you will need a separate thread, or some other form of asynchronous execution. Once you enter that while loop, no other code runs in that thread. Ever.
If I understand correcctly nothing works in your integrated version. In that case, yes you probably need to run the infinite loop on a separate thread, because your function with the infinit loop will never exit, so no other code will ever run on that thread.
You don't say what OS, but yes, set it as a low-priority thread, minimal stack size. I flash a LED in my projects, just so I can easily see if the code has reached the abort-handler yet :)
void LEDflash_task_code(void *p)
{
while (1)
{
FIO1CLR=STATUS_LED;
OSsleep(750);
FIO1SET=STATUS_LED;
OSsleep(250);
};
};
If you have access to hardware peripheral timers (any micrcontroller/microprocessor application), you should use those hardware timers, not threads nor software sleep().
void init_toggle_led (uint32_t interval)
{
setup_hardware_timer(interval);
}
void toggle_led (void)
{
if( (hardware_timer_register & flag) > 0 )
{
port = port ^ pin_mask;
}
}
main()
{
init_toggle_led(1000);
for(;;)
{
do_stuff();
toggle_led();
}
}
This was an example with polling. Alternatively, you can use hardware interrupts from the timers and toggle the port from there.
As David mentioned, you should run your LED code in a separate thread. http://www.yolinux.com/TUTORIALS/LinuxTutorialPosixThreads.html#BASICS
Once you have threads, if you want your code to be able to stop your LED from blinking, then add a flag that's checked inside the while loop at each iteration, and if it's set then break out.
#include <stdio.h>
#include <unistd.h>
#include <pthread.h>
void * toggle_led(void *ptr);
int stop=0;
int main (int argc, const char * argv[])
{
printf("Hello, World!\n");
// set up thread
pthread_t LED_thread;
char * message = "blink!";
pthread_create( &LED_thread, NULL, toggle_led, (void*) message);
// do some other work
sleep(5);
// ok you want to quit now
stop=1;
pthread_join(LED_thread, NULL);
printf("Goodbye!\n");
return 0;
}
void *toggle_led(void *ptr)
{
while (!stop)
{
printf("%s \n", (char *)ptr);
sleep(1);
}
}
I think you need to implement as a watchdog functionality. Because if you use threads then even if other threads has some issues(like deadlock), your LEDs will toggle as long as toggle_led thread works. You need to implement an toggle_led() function and call from each of other threads /functions before returning to make sure all other threads/functions are getting executed successfully without waiting continuously for some resources
I have a particular function (well, set of functions) that I want to start every 400ms. I'm not much of a C programmer, and so anything outside of the standard libraries is a mystery to me, as well as quite a bit within them.
My first thought is to use nanosleep to pause execution for 400ms in some sort of loop, but this of course doesn't take into account the execution time of the code I will be running. If I could measure it, and if it seemed fairly certain that it ran for the same approximate duration after 10 or 20 tests, I could then nanosleep() for the difference. This wouldn't be perfect, of course... but it might be close enough for a first try.
How do I measure the execution time of a C function? Or is there a better way to do this altogether, and what keywords do I need to be googling for?
You should be able to use settimer
int setitimer(int which, const struct itimerval *value,
struct itimerval *ovalue);
Just put the code that you want to execute every 400ms inside the SIGALRM handler. This way you don't need to account for the time that your code takes to run, which could potentially vary. I'm not sure what happens if the signal handler doesn't return before the next signal is generated.
An outline of what some of the code might look like is shown below.
void periodic_fuc(int signal_num)
{
...
signam(SIGALRM, periodic_func);
}
int main(...)
{
struct itimerval timerval = {0};
signal(SIGALRM, periodic_func);
...
timerval.it_interval.tv_usec = 400000;
timerval.it_value.tv_usec = 400000; // Wait 400ms for first trigger
settimer(ITIMER_REAL, &timerval, NULL);
while (!exit)
sleep(1);
return 0;
}
Take a look at gprof. It allows you to quickly recompile your code and generate information on which functions are being called and what is taking up the most time in your program.
I concur with torak about using setitimer(). However, since it's not clear if the interval is restarted when the SIGALRM handler exits, and you're really not supposed to do much work in a signal handler anyway, it's better to just have it set a flag, and do the work in the main routine:
#include <stdio.h>
#include <stdlib.h>
#include <signal.h>
#include <unistd.h>
#include <sys/time.h>
volatile sig_atomic_t wakeup = 0;
void alarm_handler(int signal_num)
{
wakeup = 1;
}
int main()
{
struct itimerval timerval = { 0 };
struct sigaction sigact = { 0 };
int finished = 0;
timerval.it_interval.tv_usec = 400000;
timerval.it_value.tv_usec = 400000;
sigact.sa_handler = alarm_handler;
sigaction(SIGALRM, &sigact, NULL);
setitimer(ITIMER_REAL, &timerval, NULL);
while (!finished)
{
/* Wait for alarm wakeup */
while (!wakeup)
pause();
wakeup = 0;
/* Code here... */
printf("(Wakeup)\n");
}
return 0;
}
You could use gettimeofday() or clock_gettime() before and after the functions to time, and then calculate the delta between the two times.
For Linux, you can use gettimeofday. Call gettimeofday at the start of the function. Run whatever you have to run. Then get the end time and figure out how much longer you have to sleep. Then call usleep for the appropriate number of microseconds.
Look at POSIX timers. Here is some documentation at HP.
You can do the same functions as with setitimer, but you also have timer_getoverrun() to let you know if you missed any timer events during your function.
The example code of section 10.6, the expected result is:
after several iterations, the static structure used by getpwnam will be corrupted, and the program will terminate with SIGSEGV signal.
But on my platform, Fedora 11, gcc (GCC) 4.4.0, the result is
[Langzi#Freedom apue]$ ./corrupt
in sig_alarm
I can see the output from sig_alarm only once, and the program seems hung up for some reason, but it does exist, and still running.
But when I try to use gdb to run the program, it seems OK, I will see the output from sig_alarm at regular intervals.
And from my manual, it said the signal handler will be set to SIG_DEF after the signal is handled, and system will not block the signal. So at the beginning of my signal handler I reset the signal handler.
Maybe I should use sigaction instead, but I only want to know the reason about the difference between normal running and gdb running.
Any advice and help will be appreciated.
following is my code:
#include "apue.h"
#include <pwd.h>
void sig_alarm(int signo);
int main()
{
struct passwd *pwdptr;
signal(SIGALRM, sig_alarm);
alarm(1);
for(;;) {
if ((pwdptr = getpwnam("Zhijin")) == NULL)
err_sys("getpwnam error");
if (strcmp("Zhijin", pwdptr->pw_name) != 0) {
printf("data corrupted, pw_name: %s\n", pwdptr->pw_name);
}
}
}
void sig_alarm(int signo)
{
signal(SIGALRM, sig_alarm);
struct passwd *rootptr;
printf("in sig_alarm\n");
if ((rootptr = getpwnam("root")) == NULL)
err_sys("getpwnam error");
alarm(1);
}
According to the standard, you're really not allowed to do much in a signal handler. All you are guaranteed to be able to do in the signal-handling function, without causing undefined behavior, is to call signal, and to assign a value to a volatile static object of the type sig_atomic_t.
The first few times I ran this program, on Ubuntu Linux, it looked like your call to alarm in the signal handler didn't work, so the loop in main just kept running after the first alarm. When I tried it later, the program ran the signal handler a few times, and then hung. All this is consistent with undefined behavior: the program fails, sometimes, and in various more or less interesting ways.
It is not uncommon for programs that have undefined behavior to work differently in the debugger. The debugger is a different environment, and your program and data could for example be laid out in memory in a different way, so errors can manifest themselves in a different way, or not at all.
I got the program to work by adding a variable:
volatile sig_atomic_t got_interrupt = 0;
And then I changed your signal handler to this very simple one:
void sig_alarm(int signo) {
got_interrupt = 1;
}
And then I inserted the actual work into the infinite loop in main:
if (got_interrupt) {
got_interrupt = 0;
signal(SIGALRM, sig_alarm);
struct passwd *rootptr;
printf("in sig_alarm\n");
if ((rootptr = getpwnam("root")) == NULL)
perror("getpwnam error");
alarm(1);
}
I think the "apue" you mention is the book "Advanced Programming in the UNIX Environment", which I don't have here, so I don't know if the purpose of this example is to show that you shouldn't mess around with things inside of a signal handler, or just that signals can cause problems by interrupting the normal work of the program.
According to the spec, the function getpwnam is not reentrant and is not guaranteed to be thread safe. Since you are accessing the structure in two different threads of control (signal handlers are effectively running in a different thread context), you are running into this issue. Whenever you have concurrent or parallel execution (as when using pthreads or when using a signal handler), you must be sure not to modify shared state (e.g. the structure owned by 'getpwnam'), and if you do, then appropriate locking/synchronization must be used.
Additionally, the signal function has been deprecated in favor of the sigaction function. In order to ensure portable behavior when registering signal handlers, you should always use the sigaction invocation.
Using the sigaction function, you can use the SA_RESETHAND flag to reset the default handler. You can also use the sigprocmask function to enable/disable the delivery of signals without modifying their handlers.
#include <stdio.h>
#include <stdlib.h>
#include <signal.h>
#include <unistd.h>
void sigalrm_handler(int);
int main()
{
signal(SIGALRM, sigalrm_handler);
alarm(3);
while(1)
{
}
return 0;
}
void sigalrm_handler(int sign)
{
printf("I am alive. Catch the sigalrm %d!\n",sign);
alarm(3);
}
For example, my code is runing in main doing nothing and every 3 seconds my program says im alive x)
I think that if you do as i done calling in the handler function alarm with value 3, the problem is resolved :)