信号量和并发编程

Question

对于家庭作业，我需要编写以下场景。这将通过使用 BACI（即 C--）的信号量来完成

有 2 间男女皆宜的洗手间，每间可容纳 4 人。由于它是男女皆宜的，因此只有同性的人可以同时在洗手间，FIFO 并不重要。我脑子里有一个基本的“算法”来处理 4 个男人和 4 个女人的 1 个洗手间。但我不知道如何编码。任何帮助将不胜感激。这是我所拥有的。

Woman:

Check to see if there are any men in the restroom. If so "wait".
If no men check to see if there are 4 people. If so "wait".
If no men and not 4 use restroom. When leaving signal there is a vacancy.
If last woman signal the men if they are waiting if not signal the woman.


Man:

check to see if there are any woman in the restroom. if so "wait"
If no woman check to see if there are 4 people. If so "wait".
If no woman and not 4 use restroom. when leaving signal there is a vacancy.
if last man signal the women if they are waiting if not signal the men.

提供了这些附加说明

• 使用随机 FOR 循环在适当的地方模拟时间的流逝。这可以通过使用延迟函数轻松完成：

  void Delay (void)
  { 
    int i;
    int DelayTime;
    DelayTime = random (DELAY);
    for (i = 0; i < DelayTime; i++):
  }
  

• 其中 const int DELAY = 10 到 100 之间的某个数字。

• 很好地打印和格式化输出，并以这样一种方式打印消息：通过读取输出，人们可以跟踪执行顺序。
• 将进程设置为永远循环，并使用控制 C（或控制中断）停止您的程序。

Answer 1

适用于 4 个卫生间的 baci 代码：

 const int Delayx = 60;
   int i;
   int Mcount,Wcount;
   binarysem x,y,z,Wsem,Msem;
   semaphore cap;
   void Delay(void)
   {
    int DelayTime;
    DelayTime = random(Delayx);
    for (i = 0; i<DelayTime; i++);
     }

 void Woman(void)
   {
     wait(z);
     wait(Wsem);
     wait(y);
     Wcount++;
     if(Wcount==1)
       { wait(Msem);  }
       signal(y);
       signal(Wsem);
       signal(z);

       wait(cap);
       cout << "A Woman has entered Restroom"<<endl;
       Delay();
       cout << "A Woman has exited Restroom"<<endl;

       signal(cap);
       wait(y);
       Wcount--;
       if(Wcount==0)
         {signal(Msem);}

        signal(y);
        }

 void Man(void)
  {
     wait(z);
     wait(Msem);
     wait(x);
     Mcount++;
     if(Mcount==1)
       { wait(Wsem);  }
       signal(x);
       signal(Msem);
       signal(z);

       wait(cap);
       cout << "A Man has entered Restroom"<<endl;
       Delay();
       cout << "A Man has exited Restroom"<<endl;

       signal(cap);
       wait(x);
       Mcount--;
       if(Mcount==0)
         {signal(Wsem);}

        signal(x);
        }


void main()
{
Mcount=0;
Wcount=0;
initialsem(x,1);
initialsem(y,1);
initialsem(z,1);
initialsem(Wsem,1);
initialsem(Msem,1);
initialsem(cap,4);
cobegin
{
    Woman(); Woman(); Woman();
    Woman(); Woman(); Woman(); 
    Woman();
    Woman(); Man();  Man();
    Man(); Man(); Man(); Man();
    Man(); Man();
}
      }

Answer 2

由于您想知道how to code your algorithm for 1 restroom，我已经在 C 中完成了。将它转换为 C-- 将是一项相当简单的任务，因为所有信号量构造看起来都非常相似。

根据你的回答，

C: sem_wait()  C--: wait()
   sem_post()       signal()
   sem_t            semaphore()
   sem_init()       initialsem() 

请记住，如上所述，我只解决了 1-restroom 的问题。由于这是家庭作业，我希望您自己将其扩展为 2-restrooms 形式。

从Readers-writers problem 到我们的“中性厕所”问题，我们使用以下全局变量：

int mcount,wcount; // count of number of men/women in restroom
sem_t x,y,z;       // semaphores for updating mcount & wcount values safely
sem_t wsem,msem;   // semaphores to block other genders' entry  
sem_t cap;         // capacity of the restroom

将这些信号量和计数器合并到man 线程函数中，

void *man(void *param)
{           
    sem_wait(&z);                
        sem_wait(&msem);        
            sem_wait(&x);
                mcount++;
                if(mcount==1)   
                { sem_wait(&wsem); } // first man in, make women wait
            sem_post(&x);
        sem_post(&msem);
    sem_post(&z);

    sem_wait(&cap);  //wait here, if over capacity

    printf("\t\tman in!\n");
    delay();
    printf("\t\t\tman out!\n");

    sem_post(&cap);  //one man has left, increase capacity

    sem_wait(&x);
        mcount--;
        if(mcount==0)
        {sem_post(&wsem);}  // no man left, signal women 
    sem_post(&x);
}

类似地，女性线程函数将mcount 替换为wcount，将msem 替换为wsem，将x 替换为y。只有z 在man 函数中保持原样，因此man 和woman 线程在同一个公共信号量上排队。 （因此，代码总是具有类似FIFO的行为，这确保了公平/非饥饿）

完整代码如下：（编译使用gcc filename -lpthread）

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

int mcount,wcount;
sem_t x,y,z,wsem,msem,cap;

void delay(void)
{
    int i;
    int delaytime;
    delaytime = random();
    for (i = 0; i<delaytime; i++);
}

void *woman(void *param)
{
    sem_wait(&z);
        sem_wait(&wsem);
            sem_wait(&y);
                wcount++;
                if(wcount==1)
                { sem_wait(&msem); }
            sem_post(&y);
        sem_post(&wsem);
    sem_post(&z);

    sem_wait(&cap);

    printf("woman in!\n");
    delay();
    printf("\twoman out!\n");

    sem_post(&cap);     

    sem_wait(&y);
        wcount--;
        if(wcount==0)
        { sem_post(&msem); }
    sem_post(&y);
}

void *man(void *param)
{           
    sem_wait(&z);
        sem_wait(&msem);
            sem_wait(&x);
                mcount++;
                if(mcount==1)
                { sem_wait(&wsem); }
            sem_post(&x);
        sem_post(&msem);
    sem_post(&z);

    sem_wait(&cap);

    printf("\t\tman in!\n");
    delay();
    printf("\t\t\tman out!\n");

    sem_post(&cap);

    sem_wait(&x);
        mcount--;
        if(mcount==0)
        {sem_post(&wsem);}
    sem_post(&x);
}

int main(void)
{
    int i;
    srandom(60);

        mcount = 0;
        wcount = 0;
        sem_init(&x,0,1);  // for sem_init, initial value is 3rd argument
        sem_init(&y,0,1);
        sem_init(&z,0,1);
        sem_init(&wsem,0,1);
        sem_init(&msem,0,1);
        sem_init(&cap,0,4);  // eg. cap initialized to 4

        pthread_t *tid;
        tid = malloc(80*sizeof(pthread_t));

    // You can use your cobegin statement here, instead of pthread_create()     
    // I have forgone the use of pthread barriers although I suppose they would nicely imitate the functionality of cobegin. 
    // This is merely to retain simplicity.

    for(i=0;i<10;i++)
    {
        pthread_create(&tid[i],NULL,woman,NULL);
    }
    for(i=10;i<20;i++)
    {     
            pthread_create(&tid[i],NULL,man,NULL);
    }
    for(i=0;i<20;i++)
    {     
            pthread_join(tid[i],NULL);
    }

    return(0);
}

在转换为 2-restrooms 形式时，请记下您需要复制哪些信号量和计数器变量以满足所有条件。快乐信号！

Answer 3

这就是我所拥有的。这允许 1 人一次进入洗手间，而不会出现僵局或饥饿。我需要有关如何制作它的帮助，以便一次可以有 4 个人在洗手间。

const int Delayx = 60;
int i;
semaphore max_capacity;
semaphore woman;
semaphore man;
semaphore mutex;

void Delay(void)
{
    int DelayTime;
    DelayTime = random(Delayx);
    for (i = 0; i<DelayTime; i++);
}

void Woman(void)
{
    wait(woman);
    wait(max_capacity);
    wait(mutex);
    cout << "A Woman has entered Restroom"<<endl;
    Delay();
    cout << "A woman has exited Restroom"<<endl;
    signal(mutex);
    signal(max_capacity);
    signal(man);
}

void Man(void)
{
    wait(man);
    wait(max_capacity);
    wait(mutex);
    cout <<"A Man has entered the Restroom"<<endl;
    Delay();
    cout << "A man has exited the Restroom"<<endl;
    signal(mutex);
    signal(max_capacity);
    signal(woman);
}

void main()
{
    initialsem(woman,1);
    initialsem(man,1);
    initialsem(max_capacity,4);
    initialsem(mutex,1);
    cobegin
    {
        Woman(); Woman(); Woman(); Woman(); Woman(); Woman(); Woman(); Woman(); Man();  Man(); Man(); Man(); Man(); Man(); Man(); Man();
    }
}