C 共享变量的多进程竞争

我正在做一个简单的生产者/消费者问题。我有一个制作人（厨师），当消费者（野蛮人）吃掉所有的食物时，他会生产食物，所以野蛮人必须等到厨师把锅装满。我不明白我的错误是什么，因为野蛮人会吃一份，但厨师不会把锅装满

以下是节目：

#include <stdio.h>
#include <stdlib.h>
#include <semaphore.h>
#include <sys/types.h>
#include <unistd.h>
#include <sys/ipc.h>
#include <sys/sem.h>
#include <sys/shm.h>
#include <sys/wait.h>
#include <fcntl.h>

int shmid, semid;
int *portions;
sem_t *mutex, *empty, *full;

char sem_1[]= "mutex";
char sem_2[]= "full";
char sem_3[]= "empty";

void clear()
{

   if (shmctl(shmid,IPC_RMID,0) == -1) perror("shmctl");
}

void producer(int num, int m)   
{
  int i;

   while(1) { 


    }    
}

void consumer(int num, int rounds)   
{

       int i;



}


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

int i;
int N, M, NROUNDS, pid;


if (argc != 4)
{
    fprintf(stderr,"insert N savages, M portions e NROUNDS\n");
    exit(1);
}

N=atoi(argv[1]);
M=atoi(argv[2]);
NROUNDS=atoi(argv[3]);




/* generate producer and consumers */




      }

  for(i=0;i<N;i++) {
    pid=wait(NULL);
    printf("Terminate process %d\n", pid);
  }   

  clear();

}

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您的使用者接收互斥锁，发出信号，表示容器为空，然后再也不会释放互斥锁，这样，生产者就永远不能接收互斥锁并填充容器，您的进程将死锁

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如果我用这个来更改它，程序不会启动：If（（*部分）==0）{sem_post（互斥）sem_post（空）SimiAuto（FULL）阅读我链接的书，你的算法可能有超过1个问题。你确定信号量是正确的同步机制吗？考虑条件变量…

./a.out 3 5 3

Savage[2] eats
Number of portions in pot: 4
Savage[1] eats
Number of portions in pot: 3
Savage[0] eats
Number of portions in pot: 2
Savage[2] eats
Number of portions in pot: 1
Savage[1] eats
Number of portions in pot: 0
Savage[0] eats
Number of portions in pot: -1
Savage[2] eats
Number of portions in pot: -2
Terminate process 10287
Savage[1] eats
Number of portions in pot: -3
Savage[0] eats
Number of portions in pot: -4
Terminate process 10285
Terminate process 10286