c - Sem_post() 无法正常工作，既不会增加信号量的值，也不会解锁附加到该信号量的进程

Question

这个问题已经有答案了:

why is the semaphore not working? (3 个回答)

已关闭 4 年前。

sem_post() 无法正常工作，既不增加信号量的值，也不阻止附加到该信号量 block 队列的进程

#include <stdio.h>
#include <semaphore.h>
#include <unistd.h>
#include <sys/types.h>
#include <sys/wait.h>

int main()
{
    sem_t pa,ch;
    sem_init(&pa,1,1);  
    sem_init(&ch,1,0);

    pid_t p1;

    p1 = fork();
    if(p1>0)    //Parent process
    {
        while(1)
        {
        sem_wait(&pa);
        printf("Parent\n");
        sem_post(&ch);
        }
    }
    else if(p1==0)
    {
        while(1)
        {
            sem_wait(&ch);
        printf("child\n");
        sem_post(&pa);
        }
    }
    else
        printf("\nError\n");

    return 0;
}

预期输出是

Parent
child
Parent
child
Parent
child
.. and so on..

实际输出是

Parent

Answer 1

引用man sem_init:

int sem_init(sem_t *sem, int pshared, unsigned int value);

If pshared is nonzero, then the semaphore is shared between processes, and should be located in a region of shared memory (see shm_open(3), mmap(2), and shmget(2)). (Since a child created by fork(2) inherits its parent's memory mappings, it can also access the semaphore.) Any process that can access the shared memory region can operate on the semaphore using sem_post(3), sem_wait(3), and so on.

您的信号量不在共享内存中，因此 fork 的子进程不会看到父进程中所做的任何更改(反之亦然)。因此两个进程都会陷入死锁。