posix信号灯故障

Question

I supposed that my program should work like this: 1) initializing unnamed semaphore with value = 0 the second value for sem_init (..) is 1 so as it said in MAN the semaphore is shared between processes 2) creating child, child waits until semaphore value becomes 1

parent process increases the value of semaphore so the child should exit now but it doesn't exit really, so that is the problem

#include <stdlib.h>
#include <stdio.h>
#include <unistd.h>
#include <semaphore.h>


pid_t child;


void child_proc(sem_t* sem) {
    sem_wait(sem);
    printf("OK\n");
}

void parent_proc(sem_t* sem) {

    sem_post(sem);
    sleep(2);

    int status;
    waitpid(child, &status, 0);
}

int main(int argc, char* argv[]) {
    sem_t sem;
    sem_init(&sem, 1, 0);

    child = fork();

    if (0 == child) {
        child_proc(&sem);
        return 0;
    }

    parent_proc(&sem);
    return 0;
}

Answer 1

The problem is that both processes have a local (not shared) copy of the semaphore structure and changes in one process won't reflect to the other process.

As the man page also says , if you want to share semaphores across processes, not only do you need to pass a non-zero value to the second argument of sem_init , but the sem_t structure also needs to exist in an area of shared memory. In your example program, it exists on the stack, which is not shared.

You can have shared memory by using a common file mapping (with mmap ) or with shm_open , notably.