Why does sem_open work with fork() without shared memory?
c, linux, semaphore
Solution
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.
Problem
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"); } ```