Question

不知道为什么消费者会做所有的工作？

我为prodcut-consumer创建了一个包含10个整数数组的信号量，数组填充了名称，并以1和0（二进制）形式返回。即使生产者正在取消信号量，也会调用消费者。

为什么会这样？

这是代码：

#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <string.h>
#include <sys/ipc.h>
#include <sys/shm.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <semaphore.h>
#include <fcntl.h>
#define SIZE 10
#define KEY 1234
int *Memory;
int i, j;
sem_t *sem;
char *name = "Hello";
int main(int argc, const char *argv[])
{
    int shmid;
    if ((shmid = shmget(KEY, sizeof(int) * SIZE, IPC_CREAT | S_IRWXU)) < 0)
    {
        perror("Error while creating shmget");
        return 1;
    }
    pid_t pid;
    sem = sem_open(name, O_CREAT, S_IRUSR | S_IWUSR, 1);
    if ((pid = fork()) != 0)
    {
        if ((shmid = shmget(KEY, sizeof(int) * SIZE, S_IRWXU)) < 0)
        {
            perror("error in shmget");
            return 1;
        }
        Memory = (int *)shmat(shmid, NULL, 0);
        if (Memory == NULL)
        {
            perror("error in shmat");
            return 1;
        }
        for (i = 0; i < 10; i++)
        {
            sem_wait(sem);
            Memory[j] = i;
            printf("Produced %i in box %i\n", i + 1, i + 1);
            sem_post(sem);
            sleep(1);
        }
        int status;
        wait(&status);
        sem_unlink(name);
        sem_destroy(sem);
        struct shmid_ds shmid_ds1;
        if (shmctl(shmid, IPC_RMID, &shmid_ds1) < 0)
        {
            perror(
                "Error in the father while executing shmctl when it was "
                "elimnating the segment of shared memory");
        }
    }
    else
    {
        if ((shmid = shmget(KEY, sizeof(int) * SIZE, S_IRWXU)) < 0)
        {
            perror("error in the producer with the shmget");
            return 1;
        }
        Memory = (int *)shmat(shmid, NULL, 0);
        if (Memory == NULL)
        {
            perror("error in the producer with the  shmat");
            return 1;
        }
        for (i = 0; i < 10; i++)
        {
            sem_wait(sem);
            Memory[i] = -1;
            printf("Consume and now it is %i in box %i\n", Memory[i], i + 1);
            sem_post(sem);
        }
    }
    return 0;
}

输出是：

Produced 1 in box 1
Consume and now it is -1 in box 1
Consume and now it is -1 in box 2
Consume and now it is -1 in box 3
Consume and now it is -1 in box 4
Consume and now it is -1 in box 5
Consume and now it is -1 in box 6
Consume and now it is -1 in box 7
Consume and now it is -1 in box 8
Consume and now it is -1 in box 9
Consume and now it is -1 in box 10
Produced 2 in box 2
Produced 3 in box 3
Produced 4 in box 4
Produced 5 in box 5
Produced 6 in box 6
Produced 7 in box 7
Produced 8 in box 8
Produced 9 in box 9
Produced 10 in box 10

Answer 1

#include <pthread.h>
#include <stdio.h>
#include <semaphore.h>
const int max = 5;
int arr[5], f = 0, r = -1;
sem_t s1, s2, sm;
void* eprod(void* pv)
{
    int i, x;
    printf("Producer Welcome\n");
    // sleep(10);
    while (1)
    {
        x = rand() % 100;
        printf("producer going to add:%d\n", x);
        sem_wait(&s2);
        sem_wait(&sm);
        // down s2 //buffer may be full
        // lock sm
        r = (r + 1) % max;
        arr[r] = x;
        sem_post(&sm);
        sem_post(&s1);
        sleep(10);
    }
    // unlock sm
    // up s1
}
void* econs(void* pv)
{
    int i, x;
    printf("Consumer Welcome\n");
    while (1)
    {
        sem_wait(&s1);
        sem_wait(&sm);
        // down s1
        // lock sm
        x = arr[f];
        f = (f + 1) % max;
        printf("Consumer removed element:%d\n", x);
        sem_post(&sm);
        sem_post(&s2);
        // unlock sm
        // up s2
        // sleep(5);
    }
}
int main()
{
    pthread_t pt1, pt2;
    sem_init(&s1, 0, 0);    // 3rd Parameter ival=1
    sem_init(&s2, 0, max);  // 3rd Parameter ival=1
    sem_init(&sm, 0, 1);    // 3rd Parameter ival=1
    pthread_create(&pt1, NULL, eprod, (void*)0);
    pthread_create(&pt2, NULL, econs, (void*)1);
    printf("Main Thread is Running\n");
    pthread_join(pt1, NULL);
    pthread_join(pt2, NULL);
    printf("Main -- - - Thanks..!\n");
    return 0;
}

希望这会有所帮助..

Answer 2

您的输出与您的程序代码一致。您的两个进程都使用信号量来访问共享内存块，因此它们通过互斥来工作，即只允许其中一个在任何给定时间使用Memory[]数组。

但是你的程序中没有任何东西会进一步限制进程的行为，因此消费者可以在没有生产者做任何事情的情况下继续进行。

您需要在程序中引入更多bookkeping来处理Memory[]数组中的保留/空闲槽，以及另外两个用于同步生产者和使用者进度的信号量。查看问题的任何标准解决方案，例如http://en.wikipedia.org/wiki/Producer%E2%80%93consumer_problem#Using_semaphores

具有信号量和叉子的产品消费者

2 个答案: