This is supposed to work in a loop (server) and delegate work/inquiry to a faulty library, here represented by the longrun() function call, to a thread with a time-out of tmax=3s. I placed synchronization vars and i am trying to wait for no more than this limit, but when longrun() hangs (run 4), it still waits the full time (7s) instead of the requested limit. Can anyone explain?
#include <unistd.h>
#include <errno.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <time.h>
#include <pthread.h>
#include <sys/time.h>
#include <iostream>
using namespace std;
string int2str(int i){
char buf[10]; // no larger int passed we hope
int end = sprintf(buf, "%d", i);
buf[end] = '';
return string(buf);
}
string longrun(int qi){
if(qi % 4 == 0) {
sleep(7);
return string("'---- to: ") + int2str(qi) + string("' (hang case)");
}
else {
sleep(1);
return string("'okay to: ") + int2str(qi) + string("'");
}
}
struct tpack_t { // thread pack
pthread_t thread;
pthread_mutex_t mutex;
pthread_cond_t go; // have a new value to run
pthread_cond_t ready; // tell main thread we're done processing
int newq; // predicate on go+ready condition for wait
int qi; // place question as int to thread: question-int
string res; // where i place the response
tpack_t();
};
tpack_t::tpack_t() {
pthread_mutex_init (&mutex, NULL);
pthread_cond_init (&go, NULL);
pthread_cond_init (&ready, NULL);
newq = 0;
}
void set_cond_time(timespec* ctp, int tmax){
timeval now;
gettimeofday(&now, NULL);
ctp->tv_nsec = now.tv_usec * 1000UL;
ctp->tv_sec = now.tv_sec + tmax; // now + max time!
printf("[m] ... set to sleep for %d sec, i hope...n", tmax);
}
void take_faulty_evasive_action(tpack_t* tpx){
// basically kill thread, clean faulty library copy (that file) and restart it
cout << "will work on it (restarting thread) soon!n";
tpx->newq = 0; // minimal action for now...
}
void* faulty_proc(void* arg){
tpack_t* tpx = (tpack_t*) arg;
while(true){
pthread_mutex_lock(&tpx->mutex);
while(tpx->newq == 0){
pthread_cond_wait(&tpx->go, &tpx->mutex);
}
printf("[t] to process : %dn", tpx->qi); fflush(stdout);
// now i have a new value in qi, process it and place the answer in... res
tpx->res = longrun(tpx->qi);
tpx->newq = 0;
pthread_mutex_unlock(&tpx->mutex);
pthread_cond_signal(&tpx->ready);
}
}
int main(int argc, char* argv[]){
cout << "n this presents the problem: idx = 4k -> hang case ...n ( challenge is to eliminate them by killing thread and restarting it )nn";
printf(" ETIMEDOUT = %d EINVAL = %d EPERM = %dnn", ETIMEDOUT, EINVAL, EPERM);
tpack_t* tpx = new tpack_t();
pthread_create(&tpx->thread, NULL, &faulty_proc, (void*) tpx);
// max wait time; more than that is a hanging indication!
int numproc = 5;
++numproc;
int tmax = 3;
timespec cond_time;
cond_time.tv_nsec = 0;
int status, expired; // for timed wait on done condition!
time_t t0 = time(NULL);
for(int i=1; i<numproc; ++i){
expired = 0;
pthread_mutex_lock(&tpx->mutex);
tpx->qi = i; // init the question
tpx->newq = 1; // ... predicate
//pthread_mutex_unlock(&tpx->mutex);
pthread_cond_signal(&tpx->go); // let it know that...
while(tpx->newq == 1){
/// ---------------------- most amazing region, timedwait waits all the way! ----------------------
set_cond_time(&cond_time, tmax); // time must be FROM NOW! (abs time, not interval)
time_t wt0 = time(NULL);
status = pthread_cond_timedwait(&tpx->ready, &tpx->mutex, &cond_time);
printf("[m] ---- t exited with status = %d (after %.2fs)n", status, difftime(time(NULL), wt0));
/// -----------------------------------------------------------------------------------------------
if (status == ETIMEDOUT){
printf("t ['t was and newq == %d]n", tpx->newq);
if(tpx->newq == 1){ // check one more time, to elim race possibility
expired = 1;
break;
}
}
else if(status != 0){
fprintf(stderr, "cond timewait for faulty to reply errored outn");
return 1;
}
}
if(expired){
take_faulty_evasive_action(tpx); // kill thread, start new one, report failure below
cout << "[m] :: interruption: default bad answer goes here for " << i << "nn";
}
else {
cout << "[m] :: end with ans: " << tpx->res << endl << endl;
}
pthread_mutex_unlock(&tpx->mutex);
}
time_t t1 = time(NULL);
printf("took %.2f sec to runn", difftime(t1, t0));
}
Used ‘g++ -pthread code.cc’ to compile under linux (ubuntu 16.04). Output is:
this presents the problem: idx = 4k -> hang case ...
( challenge is to eliminate them by killing thread and restarting it )
ETIMEDOUT = 110 EINVAL = 22 EPERM = 1
[m] ... set to sleep for 3 sec, i hope...
[t] to process : 1
[m] ---- exited with status = 0 (after 1.00s)
[m] :: end with ans: 'okay to: 1'
[m] ... set to sleep for 3 sec, i hope...
[t] to process : 2
[m] ---- exited with status = 0 (after 1.00s)
[m] :: end with ans: 'okay to: 2'
[m] ... set to sleep for 3 sec, i hope...
[t] to process : 3
[m] ---- exited with status = 0 (after 1.00s)
[m] :: end with ans: 'okay to: 3'
[m] ... set to sleep for 3 sec, i hope...
[t] to process : 4
[m] ---- exited with status = 110 (after 7.00s)
['t was and newq == 0]
[m] :: end with ans: '---- to: 4' (hang case)
[m] ... set to sleep for 3 sec, i hope...
[t] to process : 5
[m] ---- exited with status = 0 (after 1.00s)
[m] :: end with ans: 'okay to: 5'
took 11.00 sec to run
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Answer
The problem is that faulty_proc() keeps tpx->mutex locked while it calls longrun(), and the pthread_cond_timedwait() call in main() can’t return until it can re-acquire the mutex, even if the timeout expires.
If longrun() doesn’t need the mutex to be locked – and that seems to be the case – you can unlock the mutex around that call and re-lock it before setting the completion flag and signalling the condition variable.