Why does sem_open work with fork() without shared memory?

问题

This program works (I tested it), even though the semaphore is not in shared memory. Note how I create the variable once - before the fork().

On the other hand, a semaphore created with sem_init() needs to be in shared memory to work. But it's still a sem_t structure, so why doesn't it require shared memory?

Are the contents of the sem_t structure somehow different?

sem_t *s = sem_open("mysemaphore1", O_CREAT, 0600, 0);
if (fork()) {
    sleep(3);
    sem_post(s);
} else {
    sem_wait(s);
    printf("Woke\n");
}

回答1:

The semaphore created by sem_open() is a named semaphore. The basic purpose of named semaphore is to be used between unrelated processes. The semaphore created by sem_init() is an unnamed semaphore. It is light weight than the named semaphore and needs to be put in shared memory if used between related processes. If used between threads of the same process, it can be kept in global variable.

The pointer returned by the sem_open() is actually a pointer to a memory mapped by mmap() with MAP_SHARED flag set. Since such kind of memory persists across fork(), hence you are able to use the same variable in both parent and child to access the named semaphore.

来源：https://stackoverflow.com/questions/21129845/why-does-sem-open-work-with-fork-without-shared-memory