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我试图让我的程序做到这一点：

  • 接受输入：nrNodes NrWorkers
  • 3个线程（工作者）一次只能访问列表（读取），但只有1个可以写入。

  • 当 5 个节点已完成（sqrt 值）时，它应该停止并让在链表中清除所有已完成节点的更清洁线程，而不是将访问权返回给剩余的有工作的线程

    Thread1: 获取 semaphoreSlot(); 工作; 如果条件完成发送信号；直到列表被解析；

    Thread2：等待信号；干净的; 返回对thread1的访问；

    清理后：遇到问题；停止，不退出；

代码 ：

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <unistd.h>
#include <math.h>
#include <semaphore.h>

#define doneLimit 5

struct dataBlock{
    struct node *root;
    int listSize;
    int forIndex;
};

struct node { // std linked list node
    int value;
    int worker;
    struct node *next;
};

int done = 0;

sem_t sem;

pthread_mutex_t mutp;   // mutex
pthread_cond_t  condvar;   //condvar

void *deleteDoneNodes(void *n){
    pthread_cond_wait( &condvar, &mutp );
    struct node *root = n;
    struct node *it = root;
    struct node *prev = NULL;

    printf("Cleaning 1's \n");   
        do{
            if(it->value == 1){
                struct node *next = it->next;
                if (prev != NULL) {
                    prev->next = next;
                }
                if (it == root) {
                    root = next;
                }
                free(it);
                it = next;
            }
            else {
                prev = it;
                it = it->next;
            }
        }while(it !=  NULL);
        done = 0;

    pthread_exit(NULL);
}

void * worker( void *data ){
    // get list
    int wFlag;
    struct dataBlock *inData = ( struct dataBlock * ) data;
    struct node *root = inData->root;
    int forIndex = inData->forIndex;
    free(data);

    // parse
    while(1){
        if( sem_wait( &sem )  != 0 ){
            printf( " > waiting...  \n" );
        }
        struct node *it = root;

        printf("    Thread >>>  %d  ---  %lu \n", forIndex, pthread_self() );
        do{
            wFlag = 0;
            pthread_mutex_lock( &mutp );
            if( forIndex == it->worker ){
                if( it->value > 2 ){

                    while( it->value != 1 ){
                        it->value = sqrt(it->value);
                        }
                        printf("! node done\n");
                    pthread_mutex_unlock( &mutp );
                    wFlag += 1;
                    done += 1;
                    if( done == doneLimit ){ // limit 5
                        pthread_cond_signal( &condvar );
                    }
                    break;
                }
            }
            else{
                printf("...parsed done node. \n");
            }
            it = it->next;
            pthread_mutex_unlock( &mutp );
        }while(it !=  NULL);

        sem_post(&sem); 
        sleep(1); // "create" concurrancy envi. 
        if ( wFlag == 0 ){
            break;
        }
    }
    pthread_exit(NULL);
}



int main( int argc, char *argv[] ){
    if ( argc != 3 ){
        printf( "Programm must be called with \n NR of elements and NR of workers! \n " );
        exit( 1 );
    }
    int i;
    struct node *root;
    struct node *iterator;  

//prepare list for task
    int listSize = atoi(argv[1]);
    int nrWorkers = atoi(argv[2]);
    root = malloc(sizeof( struct node) );

    root->value = rand() % 100;
    root->worker = 0;
    iterator = root;

    for( i=1; i<listSize; i++ ){
        iterator->next = malloc(sizeof(struct node));
        iterator = iterator->next;
        iterator->value = rand() % 100;
        iterator->worker = i % nrWorkers;
        printf("node #%d worker: %d  value: %d\n", i, iterator->worker,iterator->value);
    }
    iterator->next = NULL;
    printf("? List got populated\n");
// init semaphore > keeps max 3 threads working over the list

    if( sem_init(&sem,0,3) < 0){
      perror("semaphore initilization");
      exit(0);
    }

// Create all threads to parse the link list
    int ret;    
    pthread_mutex_init( &mutp,NULL );
    pthread_cond_init( &condvar, NULL );

    pthread_t w_thread;
    pthread_t* w_threads = malloc(nrWorkers * sizeof(w_thread));

    for( i=0; i < nrWorkers; i++ ){         
        struct dataBlock *data = malloc(sizeof(struct dataBlock));
        data->root = root;
        data->listSize = listSize;
        data->forIndex = i;
        ret = pthread_create ( &w_threads[i], NULL, worker, (void *) data );
        if( ret ) {
            perror("Worker creation fail \n");
            exit(2);    
        }   
    } 
// Create Cleaning thread
    pthread_t c_thread;
    ret = pthread_create(&c_thread, NULL, deleteDoneNodes, (void *) root);
    if ( ret ){
        printf("Cleaner cration fail \n");
    }

// Join threads
    for ( i = 0; i < nrWorkers; i++ ){
        pthread_join(w_threads[i],NULL);
    }

    iterator = root;
    for ( i = 0; i < listSize; i++){
        printf("val: %d  worker: %d _  \n", iterator->value, iterator->worker);
        iterator = iterator->next;
    }

    free( root );
    pthread_mutex_destroy( &mutp );
    pthread_cond_destroy( &condvar );
    sem_destroy( &sem );

    return 0;
}

PS：它有效 ./s 1 1 ./s 1 4 ./s 4 4 ./s 4 1

调用清洁器后失败：当 x >= 5 时，它在 ./sxn 处失败

4

1 回答 1

2

至少您没有正确处理条件变量和互斥锁的使用。

互斥锁应在pthread_cond_wait()调用后者之前由线程调用锁定。

逐字逐句man pthread_cond_wait：

pthread_cond_wait 以原子方式解锁互斥锁（根据 pthread_unlock_mutex）并等待条件变量 cond 发出信号。线程执行被挂起并且在条件变量发出信号之前不会消耗任何 CPU 时间。调用线程必须在进入 pthread_cond_wait 时锁定互斥锁。在返回调用线程之前，pthread_cond_wait 重新获取互斥锁（根据 pthread_lock_mutex）。

此外，互斥锁在从调用返回后被锁定，pthread_cond_wait()并且必须在某个时候解锁。

于 2012-12-29T14:12:42.047 回答