linux - 为什么 pthread_cond_timedwait 在指定的时间限制后不触发？

Question

这应该在一个循环(服务器)中工作，并将工作/查询委托(delegate)给一个有故障的库，这里由 longrun() 函数调用表示，给一个超时时间为 tmax=3s 的线程。我放置了同步变量，我试图等待不超过这个限制，但是当 longrun() 挂起(运行 4)时，它仍然等待全部时间(7 秒)而不是请求的限制。谁能解释一下？

#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] = '\0';
    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 : %d\n", 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 )\n\n";
    printf("    ETIMEDOUT = %d   EINVAL = %d  EPERM = %d\n\n", 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 out\n");
                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  << "\n\n";
        }
        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 run\n", difftime(t1, t0));
}

使用 'g++ -pthread code.cc' 在 linux (ubuntu 16.04) 下编译。输出是:

    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

Answer 1

问题是 faulty_proc() 在调用 longrun() 时保持 tpx->mutex 锁定，而 pthread_cond_timedwait () 在 main() 中调用无法返回，直到它可以重新获取互斥体，即使超时到期也是如此。

如果 longrun() 不需要锁定互斥量 - 似乎是这种情况 - 您可以解锁该调用周围的互斥量并在设置完成标志之前重新锁定它并向条件变量发送信号。