c - pthread执行陷入无限循环

Question

我正在尝试的是:

• 创建了四个线程(main创建了3个线程，线程3创建了线程4)
• 一个共享内存对象在所有线程之间共享。
• 另一个共享内存对象在线程二和线程四之间共享。
• 线程 4 等待来自线程 2 的信号，直到创建共享内存对象。
• 所有这些都是互斥的。

但是我的程序陷入了无限循环。需要解决方案的帮助。

下面是我的源代码:

#include <assert.h>
#include <stdlib.h>
#include <pthread.h>
#include <unistd.h>   /* for ftruncate */
#include <sys/mman.h> /* for shm_ and mmap */
#include <sys/stat.h> /* For mode constants */
#include <fcntl.h>    /* For O_* constants */

pthread_t T1, T2, T3, T4;

pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
pthread_mutex_t mutex2 = PTHREAD_MUTEX_INITIALIZER;
pthread_mutex_t mutex3 = PTHREAD_MUTEX_INITIALIZER;
pthread_cond_t cond_one;

int fd;
int *shared_heap;
int *shared_heap2;
int *shared_heap3;
int counter = 0;

//thread one creator func
// *argv is the shared mem obj which is passed while thread is created
void* task1(void *argv) {
    int *var = (int*) argv;
    pthread_mutex_lock(&mutex);
    *var += 1;
     pthread_mutex_unlock(&mutex);
    return NULL;
} 

//thread two creator func
// *argv is the shared mem obj which is passed while thread is created
void* task2(void *argv) {
    int *var = (int*) argv;
    pthread_mutex_lock(&mutex);
    *var += 1;

    pthread_mutex_unlock(&mutex);
    //another mutex to create another shared mem obj
    pthread_mutex_lock(&mutex2);
    shared_heap2 = (int *) mmap(NULL, sizeof(int), PROT_READ | PROT_WRITE,MAP_SHARED, fd, 0);
    assert(shared_heap2);
    counter++;
    //signal 
    if (counter > 0) {
     pthread_cond_signal(&cond_one);
        printf("signal is sent \n");
   }
   pthread_mutex_unlock(&mutex2);

  return NULL;
  }

 //thread four creator func
 //created from thread three
 // *argv is the shared mem obj which is passed while thread is created
 void* task4(void *argv) {
    int *var = (int*) argv;
    pthread_mutex_lock(&mutex);
    *var += 1;
    pthread_mutex_unlock(&mutex);

    pthread_mutex_lock(&mutex2);

    //waiting for signal from thread two
    while (counter > 0) {
    pthread_cond_wait(&cond_one, &mutex2);
    printf("waiting for signal. \n");
 }

    *shared_heap2 = 9;
    pthread_mutex_unlock(&mutex2);


    return NULL;  
}

////thread three creator func
void* task3(void *argv) {
     int *var = (int*) argv;
     pthread_mutex_lock(&mutex);
     *var += 1;
    pthread_mutex_unlock(&mutex);

    //thread four is create from here
     assert(pthread_create(&T4, NULL, &task4, var) == 0);
    assert(pthread_join(T4, NULL) == 0);
     return NULL;
}

int main(void) {

    pthread_cond_init(&cond_one, NULL);
     fd = shm_open("test_shared_var_heap_local", O_CREAT | O_RDWR,S_IRUSR | S_IWUSR);
    assert(fd != -1);
    assert(ftruncate(fd, sizeof(int)) == 0);
    shared_heap = (int *) mmap(NULL, sizeof(int), PROT_READ | PROT_WRITE,MAP_SHARED, fd, 0);

    assert(shared_heap);

    printf("main \n");
    //assert(shared_heap);

    assert(pthread_create(&T1, NULL, &task1, shared_heap) == 0);

    assert(pthread_create(&T2, NULL, &task2, shared_heap) == 0);

    assert(pthread_create(&T3, NULL, &task3, shared_heap) == 0);
    printf("three \n");

    assert(pthread_join(T1, NULL) == 0);
    assert(pthread_join(T3, NULL) == 0);
    assert(pthread_join(T2, NULL) == 0);

     return 0;
}

Answer 1

But my program falls in infinite loop.

你愿意

while (counter > 0) {
  pthread_cond_wait(&cond_one, &mutex2);
  printf("waiting for signal. \n");
}

但是计数器仅在任务2中设置为1一次，没有理由退出该时间

无论如何，这不是唯一的问题，在 task2 下 mutext2 counter 设置为 1 并发送信号，所以

• 第一种可能task4在设置为1之前已经完成，信号没用

• else task2 运行速度更快，并且在 task4 之前获取 mutex2，因此当 task4 将get mutex2 信号已发送，但信号未缓冲，因此它永远不会被 task4 接收

如果您想确保 task2 中受 mutex2 保护的代码在 task4 完成受 mutex2 保护的代码之前执行:

• counter 仍初始化为 0:int counter = 0;
• 在 task4 中，当 mutex2 被获取时，只需将 counter 设置为 1 并删除无用的测试 if (counter > 0) 在解锁 mutext2 之前的所有情况下发送信号
• 在 task2 中，当 mutext2 被获取时，将 while (counter > 0) { 替换为 if (counter == 0) {

这样:

• 如果task4在task2之前获取互斥体2，因为计数器仍然是0task4等待信号(解锁mutext2)，task2可以获取mutext2并发送信号并解锁mutext2并且完成后，task4 收到信号(锁定 mutext2)，然后解锁 mutext2 并完成
• 如果task2在task4之前获取mutext2，则将counter设置为1并解锁mutext2 em> 并完成，task4 可以获取 mutext2 并且不等待信号，因为 counter 为 1，因此解锁 mutext2> 并完成