如何使用信号同步进程 [英] How to sync processes using semaphore

问题描述

假设我有3个进程，包括一个父进程，我必须按P3，P1，P2的顺序执行我的程序.伙计们，请帮我如何从过程P3开始计算.

我需要的出场次数为{0,1,2,3,4,5，.. max}

作为参考，我的代码快照为:-

 #define SEM_NAME "//test.mutex"
//#define SEM_NAME2 "//test2.mutex"

int main(int argc, char const *argv[]) {
  int max = 0, i =0;
  sem_t *sem;
  sem_t *sem2;
  pid_t  pid, pid2;
  sem = sem_open(SEM_NAME, O_CREAT, O_RDWR, 1);
  sem_unlink(SEM_NAME);
  if (sem==SEM_FAILED) {
    printf("%s sem_open failed!", SEM_NAME);
    return (-1);
  }
  // sem2 = sem_open(SEM_NAME2, O_CREAT, O_RDWR, 0);
  // sem_unlink(SEM_NAME2);
  // if (sem2==SEM_FAILED) {
  //   printf("%s sem_open failed!", SEM_NAME2);
  //   return (-1);
  // }
  printf("Enter the maximum number\n");
  scanf("%d", &max);
  pid = fork();
  if(pid == 0)
  {
    i = 2;
    pid2 = fork();
    if(pid2 == 0)
    {
      i = 0;
    }
    else
    {
      sleep(2);
    }
  }
  else
  {
    i = 1;
    sleep(1);
  }
  //do
  {
    sem_wait(sem);
    for (; i <= max;) {
      printf("pid %d done and value is %d\n", getpid(),i);
      i = i + 3;
    }
    sem_post(sem);
  } //while(i <= max);
  wait(NULL);
  return 0;
}

当我运行程序时，我得到以下输出 {0,3，，6,1,4,7,2,5,8}

我需要一种方法，即第一个进程应打印一个数字，然后让其他进程打印其编号，最后一个进程应打印一个数字.

我需要一种方法，使每个方法依次到达.

希望我对这个问题很清楚

解决方案

以下是如何获得订单的伪代码 P3，P1，P2

//semaphores for P1, P2, P3 respectively
semaphore s1=0;
semaphore s2=0;
semaphore s3=1;     //coz P3 needs to go first 


//for p3
wait(s3)
  //do p3 here
signal(s1)          //signal for P1 to start


//for p1
wait(s1)
  //do p1
signal(s2)           // signal to start P2


//for p2
wait(s2)
   //do p2
signal(s3)            //signal to start p3 again if that is required or you can omit this if not required

let's say I have 3 processes including a parent process I have to execute I program in sequence of P3,P1,P2. Guys please help me how I can start the computation from process P3.

I need the out as {0,1,2,3,4,5,.. max}

For the reference my code snapshot is :-

 #define SEM_NAME "//test.mutex"
//#define SEM_NAME2 "//test2.mutex"

int main(int argc, char const *argv[]) {
  int max = 0, i =0;
  sem_t *sem;
  sem_t *sem2;
  pid_t  pid, pid2;
  sem = sem_open(SEM_NAME, O_CREAT, O_RDWR, 1);
  sem_unlink(SEM_NAME);
  if (sem==SEM_FAILED) {
    printf("%s sem_open failed!", SEM_NAME);
    return (-1);
  }
  // sem2 = sem_open(SEM_NAME2, O_CREAT, O_RDWR, 0);
  // sem_unlink(SEM_NAME2);
  // if (sem2==SEM_FAILED) {
  //   printf("%s sem_open failed!", SEM_NAME2);
  //   return (-1);
  // }
  printf("Enter the maximum number\n");
  scanf("%d", &max);
  pid = fork();
  if(pid == 0)
  {
    i = 2;
    pid2 = fork();
    if(pid2 == 0)
    {
      i = 0;
    }
    else
    {
      sleep(2);
    }
  }
  else
  {
    i = 1;
    sleep(1);
  }
  //do
  {
    sem_wait(sem);
    for (; i <= max;) {
      printf("pid %d done and value is %d\n", getpid(),i);
      i = i + 3;
    }
    sem_post(sem);
  } //while(i <= max);
  wait(NULL);
  return 0;
}

when I run the program I get following output {0,3,,6,1,4,7,2,5,8}

I need a way I which first process should print a number and it should let other process to print his number and at the last third should print.

I need a way that each and get there turn sequentially.

Hope I am clear with the question

解决方案

Here's pseudo code of how you can achieve order P3, P1, P2

//semaphores for P1, P2, P3 respectively
semaphore s1=0;
semaphore s2=0;
semaphore s3=1;     //coz P3 needs to go first 


//for p3
wait(s3)
  //do p3 here
signal(s1)          //signal for P1 to start


//for p1
wait(s1)
  //do p1
signal(s2)           // signal to start P2


//for p2
wait(s2)
   //do p2
signal(s3)            //signal to start p3 again if that is required or you can omit this if not required

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