Question

我需要编写一个创建N个子进程的程序，并且每个进程都会向共享内存变量添加一个。我的想法是使用信号量和共享内存，但进程不是在等待彼此，共享内存变量也不能正常工作。

mydefs.h

#ifndef __MYDEFS__H__
#define __MYDEFS__H__
// Includes
#include <stdio.h>
#include <unistd.h>
#include <semaphore.h>
#include <stdlib.h>
#include <signal.h>
#include <errno.h>
#include <memory.h>
#include <sys/ipc.h>
#include <sys/msg.h>
#include <sys/types.h>
#include <sys/shm.h>
#endif // __MYDEFS__H__

的main.c

#include "mydefs.h"
#define PROC_COUNT 3
#define INITAL_MARKER_VALUE 0
#define PID_LEN 32

char mypid[PID_LEN];

int main()
{
    int i, shm_id;
    sem_t mutex;
    if(sem_init(&mutex,1,1) < 0)
    {
        perror("semaphore initilization");
        exit(0);
    }
    shm_id = shmget(IPC_PRIVATE, 4*sizeof(int), IPC_CREAT | 0666);
    if (shm_id < 0) {
         printf("shmget error\n");
    }
    int *shmpointer = shmat(shm_id,0,0);
    memset(mypid, 0, sizeof(mypid));
    sprintf(mypid, "%06d", getpid());

    for(i = 0; i < PROC_COUNT; i++)
    {
        if (fork() == 0)
        {
            while(sem_wait(&mutex)!=0);
            execl("slaveproc", "slaveproc", mypid, (char *)0);
            shmpointer += 1;
            sem_post(&mutex);
            perror("\n Can't exec slave program. Cause ");
            exit(1);
        }
    }
    sleep(1);
    printf("%d\n", *shmpointer);
    return 0;
}

slaveproc.c

#include "mydefs.h"

int marker; // Marker value

int main(int argc, char *argv[])
{
    master_pid = atoi(argv[1]);
    printf("\n --------------------------------------");
    printf("\n I'm the slave proc!");
    printf("\n My pid: %d", getpid());
    printf("\n My master's pid: %d", master_pid);
    printf("\n --------------------------------------");
    for(;;) pause();
        return 0;
}

Answer 1

问题（或至少＆＃34;问题＆＃34;）是mutex不在共享内存中：它在堆栈上分配。当您fork()时，新流程将与旧流程完全分开，因此在一个流程上调用sem_wait(&mutex)不会影响其他流程mutex。

您应该将mutex放在共享内存中：

int main()
{
    int i, shm_id;
    shm_id = shmget(IPC_PRIVATE, sizeof(sem_t) + 4*sizeof(int), IPC_CREAT | 0666);
    if (shm_id < 0) {
         printf("shmget error\n");
    }
    int *shmpointer = shmat(shm_id,0,0);
    sem_t *mutex = shmpointer;
    shmpointer = (void*)shmpointer + sizeof(sem_t);
    if(sem_init(mutex,1,1) < 0)
    {
        perror("semaphore initilization");
        exit(0);
    }        
    memset(mypid, 0, sizeof(mypid));
    sprintf(mypid, "%06d", getpid());

    for(i = 0; i < PROC_COUNT; i++)
    {
        if (fork() == 0)
        {
            while(sem_wait(mutex)!=0);
            execl("slaveproc", "slaveproc", mypid, (char *)0);
            shmpointer += 1;
            sem_post(mutex);
            perror("\n Can't exec slave program. Cause ");
            exit(1);
        }
    }
    sleep(1);
    printf("%d\n", *shmpointer);
    return 0;
}

你也永远不会在shmpointer写一些内容（也许你的意思是(*shmpointer) += 1？），但我会让你自己解决这个问题。

在Linux下使用信号量和共享内存

1 个答案: