problem using pthread_cond_signal and pthread_cond_timedwait

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01-09-2024

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Debugging pthread_cond_timedwait Timeouts: A Stack Overflow Case Study

This article explores a common issue encountered when using pthread_cond_timedwait in multithreaded programs. We'll analyze a code snippet from a Stack Overflow question and provide insights to understand and resolve the problem.

The Problem

The provided code implements two threads (Thread 1 and Thread 2) that use condition variables (cond1 and cond2) and mutexes (mutex1 and mutex2) to signal each other. The program intends to run indefinitely, with each thread waiting on a specific condition and signaling the other thread when a task is complete. However, the code intermittently experiences timeouts within pthread_cond_timedwait, leading to premature termination.

Analysis

The root cause of the timeout issue lies in the incorrect use of mutexes and condition variables. Let's break down the problematic areas:

• Race Condition: Both threads attempt to acquire the same mutex (mutex2) when signaling the other thread. This creates a race condition where one thread might acquire the mutex and signal the condition variable before the other thread has entered the pthread_cond_timedwait function. Consequently, the other thread might miss the signal and eventually time out.
• Unnecessary Mutex Acquisition: Thread 1 acquires mutex1 before calling pthread_cond_timedwait on cond2, which is unnecessary. The mutex is solely used to protect cond1, not cond2. Similarly, Thread 2 acquires mutex2 before calling pthread_cond_timedwait on cond1, which is also unnecessary.

Solution

To address these issues, we need to modify the code as follows:

1. Remove unnecessary mutex acquisition: Remove the mutex lock and unlock calls before entering pthread_cond_timedwait if the mutex is not related to the condition variable being waited on.
2. Ensure mutual exclusion for signaling: Ensure that the mutex associated with a condition variable is acquired before signaling the condition and released afterward.

Here is the corrected code:

#include <errno.h>
#include <pthread.h>
#include <signal.h>
#include <stdbool.h>
#include <stdint.h>
#include <stdio.h>
#include <sys/time.h>
#include <time.h>
#include <unistd.h>

pthread_mutex_t mutex1;
pthread_cond_t cond1;
pthread_mutex_t mutex2;
pthread_cond_t cond2;
bool run = true;

void signal_handler(int sig_num) { run = false; }

void* thread1(void* arg) {
    struct timespec ts;
    struct timeval now;
    uint32_t runs = 0;

    while (run) {
        // Wait for signal from Thread 2
        pthread_mutex_lock(&mutex2);  // Acquire mutex associated with cond2
        int result = pthread_cond_timedwait(&cond2, &mutex2, &ts);
        if (result == ETIMEDOUT) {
            printf("Thread 1: Wait timed out\n");
            run = false;
        } else if (result == 0) {
            // Signal Thread 2
            pthread_mutex_lock(&mutex1);  // Acquire mutex associated with cond1
            if (pthread_cond_signal(&cond1) != 0) {
                printf("Failed to signal cond1\n");
            }
            pthread_mutex_unlock(&mutex1);  // Release mutex
            runs++;
        }
        pthread_mutex_unlock(&mutex2);  // Release mutex
    }

    printf("T1: runs=%d\n", runs);
    return NULL;
}

void* thread2(void* arg) {
    struct timespec ts;
    struct timeval now;
    uint32_t runs = 0;

    while (run) {
        // Signal Thread 1
        pthread_mutex_lock(&mutex2);  // Acquire mutex associated with cond2
        if (pthread_cond_signal(&cond2) != 0) {
            printf("Failed to signal cond2\n");
        }
        pthread_mutex_unlock(&mutex2);  // Release mutex

        // Wait for signal from Thread 1
        pthread_mutex_lock(&mutex1);  // Acquire mutex associated with cond1
        int result = pthread_cond_timedwait(&cond1, &mutex1, &ts);
        if (result == ETIMEDOUT) {
            printf("Thread 2: Wait timed out\n");
            run = false;
        } else {
            runs++;
        }
        pthread_mutex_unlock(&mutex1);  // Release mutex
    }

    printf("T2: runs=%d\n", runs);

    return NULL;
}

int main() {
    pthread_t tid1, tid2;

    signal(SIGINT, signal_handler);
    signal(SIGTERM, signal_handler);

    pthread_mutex_init(&mutex1, NULL);
    pthread_cond_init(&cond1, NULL);
    pthread_mutex_init(&mutex2, NULL);
    pthread_cond_init(&cond2, NULL);

    pthread_create(&tid1, NULL, thread1, NULL);
    pthread_create(&tid2, NULL, thread2, NULL);

    pthread_join(tid1, NULL);
    pthread_join(tid2, NULL);

    pthread_mutex_destroy(&mutex1);
    pthread_cond_destroy(&cond1);
    pthread_mutex_destroy(&mutex2);
    pthread_cond_destroy(&cond2);

    return 0;
}

Additional Considerations

1. Timeout Value: The timeout value set in pthread_cond_timedwait should be appropriate for the expected response time. If the timeout is too short, the thread might time out unnecessarily.
2. Spurious Wakeups: Be aware that pthread_cond_timedwait can return spuriously even if a condition variable has not been signaled. This is a potential source of errors if not handled correctly.

Conclusion

This analysis demonstrates a common pitfall when using pthread_cond_timedwait in multithreaded programs. Understanding the principles of mutexes, condition variables, and proper thread synchronization is crucial for robust and reliable code. By correctly managing mutexes and ensuring proper signaling, developers can avoid timeouts and create predictable multithreaded applications. Remember to consult reliable sources and code examples to understand the intricacies of thread synchronization mechanisms.