信号量和并发编程 [英] Semaphores and concurrent programming
问题描述
有关家庭作业，我需要编程以下情形。这是要通过使用BACI（这是C--）信号量来完成
For a homework assignment i need to program the following scenario. This is going to be done using semaphores using BACI (which is C--)
有2有可各容纳4人的中性厕所。既然是男女皆宜的只有同性别的人可以在同一时间在洗手间和FIFO并不重要。我在我的脑袋基本的算法来处理4男4女1洗手间。但我不知道如何code这一点。任何帮助将大大AP preciated。下面是我。
There are 2 unisex restrooms that can hold 4 people each. Since it is unisex only people of the same sex can be in the restroom at the same time and FIFO is not important. I have the basic "algorithm" in my head to handle 4 men and 4 woman for 1 restroom. But i don't know how to code this. Any help would be greatly appreciated. Here is what I have.
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.
中提供这些额外的说明
These additional instructions were provided
-
使用随机FOR循环来模拟在适当的地方时间的流逝。这可以通过使用一延迟函数很容易地完成：
Use random FOR loops to simulate the passage of time in the appropriate places. This can be easily done by using a Delay function:
void Delay (void)
{
int i;
int DelayTime;
DelayTime = random (DELAY);
for (i = 0; i < DelayTime; i++):
}
其中const int的DELAY = 10〜100一定数量
where const int DELAY = some number from 10 to 100.
推荐答案
既然你想知道如何code的1厕所算法，我已经做了所以在C.这将是一个相当简单的任务，将其转换成C--，因为所有的信号灯结构显得颇为相似。
Since you want to know how to code your algorithm for 1 restroom, I have done so in C. It will be a fairly simple task to convert it into C--, as all the semaphore constructs appear quite similar.
从什么我可以做你的答案，
From what I could make of your answer,
C: sem_wait() C--: wait()
sem_post() signal()
sem_t semaphore()
sem_init() initialsem()
记住，如前所述，我只制定出了问题对于 1洗手间。由于这是功课，我希望你把它扩展到在 2洗手间形成自己。
Bear in mind, as stated, I have worked out the problem for 1-restroom only. Since this is homework, I expect you to expand it into the 2-restrooms form yourself.
工作一个人的方式从读者，作家的问题，我们的中性厕所问题，我们使用以下全局变量：
Working one's way from the Readers-writers problem to our "Unisex Restroom" problem, we make use of the following global variables:
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＆放大器;柜台到人线程函数，
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);
}
同样，女人线程函数，替代品 mcount 与 wcount ， MSEM 与 WSEM 和 X 与是。只有以Z 仍是在人的功能，这样既人＆安培; 女人线程排队，在相同的公共信号。 （由于这一点，code总是有 FIFO样的行为，从而确保公平性/非饥饿）
Similarly, the woman thread function, substitutes mcount with wcount, msem with wsem, and x with y. Only z remains as is in the man function, so that both man & woman threads queue up on the same common semaphore. (Due to this, the code invariably has FIFO-like behaviour, which ensures fairness/non-starvation)
完整code如下：（编译，使用海湾合作委员会的文件名-lpthread ）
The complete code is as follows: (To compile, use 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);
}
在转换成 2洗手间的形式，记下该信号灯和放大器;计数器变量，你将需要复制，以满足所有条件。快乐的 semaphoring 的！
While converting into the 2-restrooms form, make note of which semaphores & counter variables you would need to duplicate to satisfy all the conditions. Happy semaphoring!
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