ipc - 使用 System V 信号量时的挑战和奇怪的经历

Question

系统:

Linux ubuntu 3.2.0-57-generic-pae i686 GNU/Linux

问题:

我想利用信号量来实现进程原子性

下面的程序只是将一些数据写入 FIFO，而其他一些进程也尝试写入同一个 FIFO，但下面首先获取信号量的进程必须完成其写入并为其他进程释放 key

header.h

#include<stdio.h>
#include<stdlib.h>
#include<unistd.h>
#include<sys/types.h>
//#include<sys/ipc.h>
//#include<sys/sem.h>
#include<fcntl.h>
#define RES_FIFO 1234
#include<linux/sem.h>
typedef struct
{
pid_t   pid;
int     op1;
int     op2;
char    oper;
}MATH;
int result;
int ret;

流程:

#include"header.h"
int
main(void)
{
 MATH request;
 int fd,KK;
 if(access("TEMP",F_OK)== -1)
    mkfifo("TEMP",0666);
 fd=open("TEMP",O_WRONLY,0666);
 int i=0;
 union semun a;
 struct sembuf sops[2]={{0,-1,SEM_UNDO},{0,1,SEM_UNDO}};
 //1.Semaphore Creation
 KK=semget(RES_FIFO,1,IPC_CREAT);//semaphore set having one semaphore is created and kernel key will be returned
 if(KK==-1)
 {
 perror("Semget Error in P1\n");
 exit(1);
 }
 //2.Semaphore Control
 a.val=1;
 ret= semctl(KK,0,a);
 if(ret==-1)
 {
  perror("SemCtl Error in P1\n");
  exit(1);
 }
 //3.Semaphore Operations
 ret=semop(KK,&sops[0],1);
 if(ret==-1)
 {
  perror("Semop Error in decrementing at P1\n");
  exit(1);
 }


 for(i=0;i<3;i++)
 {
  request.op1=i;
  request.op2=i;
  request.oper='+';
  request.pid=getpid();
  write(fd,&request,sizeof(request));
 // sleep(1);
 }

 ret=semop(KK,&sops[1],2);
 if(ret==-1)
 {
  perror("Semop Error in decrementing at P1\n");
  exit(1);
 }
return 1;
}

但问题是在信号量操作semop之后，我被卡住了，进程处于空闲状态，什么也不做，并进入休眠状态。那么我的代码有什么问题吗？

最新编辑 1:

根据相关建议，我更新了代码行

ret= semctl(KK,0,SETVAL,a);//command updated as SETVAL

并且还取消注释该行

sleep(1);

所以我的process1，process2，process3有相同的代码，这里我忘了提到从fifo读取并在控制台上打印的阅读器代码

所以我在控制台上运行这样的脚本

./reader&
./p1&
./p2&
./p3&

reader.c

#include"header.h"
void
main(void)
{
 MATH request;
 int fd;
 int ret;
 int KK;
 if(access("TEMP",F_OK)== -1)
         mkfifo("TEMP",0666);
 fd=open("TEMP",O_RDONLY,0666);
sleep(3);
printf("Server Read Started\n");
 while(1)
 {
 if( (ret=read(fd,&request,sizeof(request))) == -1)
 {
  perror("Read in Server Failed\n");
  exit(1);
 }
 else if(ret==0)
  continue;
 printf("Read from PID : %d about %d %c %d \n",request.pid,request.op1,request.oper,request.op2);

 }
printf("Read Complete\n");
}

根据我的目标，我希望输出是这样的

Server Read Started
Read from PID : 3673 about 0 + 0 
Read from PID : 3675 about 1 + 1 
Read from PID : 3674 about 2 + 2 
Read from PID : 3673 about 0 - 0 
Read from PID : 3675 about 1 - 1 
Read from PID : 3674 about 2 - 2 
Read from PID : 3673 about 0 * 0 
Read from PID : 3675 about 1 * 1 
Read from PID : 3674 about 2 * 2 

但实际观察到的输出是这样的

Server Read Started
Read from PID : 3673 about 0 + 0 
Read from PID : 3675 about 0 * 0 
Read from PID : 3674 about 0 - 0 
Read from PID : 3673 about 1 + 1 
Read from PID : 3675 about 1 * 1 
Read from PID : 3674 about 1 - 1 
Read from PID : 3673 about 2 + 2 
Read from PID : 3675 about 2 * 2 
Read from PID : 3674 about 2 - 2 

我认为我成功地清楚地解释了我的问题。任何进一步的编辑表示赞赏

Answer 1

Bug 就在代码中

//2.Semaphore Control
 a.val=1;
 ret= semctl(KK,0,a);
 if(ret==-1)
 {
  perror("SemCtl Error in P1\n");
  exit(1);
 }

我将它放置在所有三个进程中，它们是彼此的副本，当包含相同代码的 process2 开始执行已被 process1 递减的信号量时，它再次被重新初始化，这破坏了 3 个进程的原子性