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我想将一组未命名的信号量写入共享内存，这样我就可以在分叉的进程中访问它们。这是我到目前为止所拥有的：

int main(int argc, char *argv[]){

int num_process = atoi(argv[1]);
int i = 0;

int ppid = getpid();
void *addr;

int numsems = 512;
int object_size = numsems * sizeof(sem_t);

printf("declaring semaphores\n");
sem_t *sem[numsems];

//create shared memory and check that it worked
printf("opening shared memory\n");
int shmem_fd = shm_open("/my_shmem", O_CREAT | O_RDWR, S_IRUSR | S_IWUSR);

if(shmem_fd == -1){
    perror("Can't open shmem object");
    exit(-1);
}

//truncate memory object
if(ftruncate(shmem_fd, object_size) == -1){
    perror("failed to resize shmem object");
    exit(-1);
}

addr = mmap(NULL, object_size, PROT_READ | PROT_WRITE, MAP_SHARED, shmem_fd, 0);

if(MAP_FAILED == addr){
    perror("Map failed");
    exit(-1);
}

//store the semaphores at the start
printf("initializing semaphores\n");

for(i = 0; i<numsems; i++){
    sem[i] = addr + i*sizeof(sem_t);
    if(sem_init(sem[i], 1, 0) == -1){
            perror("sem_init failed");
            exit(-1);
    }
}

//create children
while((getpid() == ppid) && (i < num_process)){


    switch(fork()){
    case -1:
    printf("fork %d failed\n", i);
    break;

    case 0:
    //child process
    printf("child created\n");
    child_program( sem, numsems);
    printf("child done\n");
    break;

    default: //parent continues on to create next child
    break;

    }
i++;
}
}

哪个编译得很好，但是当我尝试运行它时，当它到达sem_t *sem[numsems]零件时会出现分段错误。我在别处读到你不应该创建一个指向信号量的指针，但是当我没有创建并尝试时，&sem = addr我得到了一个关于需要左值的错误。任何帮助将非常感激。

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1 回答 1

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首先，为什么不在父进程中创建（和映射）共享内存段，然后在子进程中打开呢？

而且，与其拥有一个指针数组，不如简单地映射整个数组，而不使用指针：

家长：

sem_t *sem;

shm_open("/my_shmem", O_CREAT | O_RDWR ...);

sem = mmap(NULL, object_size, ...);

孩子：

semt_t *sem;

shm_open("/my_shmem", O_RDWR, 0);

sem = mmap(NULL, object_size, ...);
于 2012-11-19T02:56:31.347 回答