Question

上个学期我在C中学习了OS实习，其中第一个项目涉及制作一个线程包，然后编写一个多生产者 - 消费者程序来演示功能。然而，在获得评分反馈之后，我失去了“信号量的使用是巧妙的错误”和“程序假设抢占（例如使用产量来改变控制）”（我们从一个非抢占式线程包开始然后稍后添加抢占）。请注意，注释和示例相互矛盾。我相信它也不会假设，并且可以在两种环境中工作。）

这已经困扰了我很长一段时间 - 课程工作人员有点不知所措，所以我不能问他们这个学期有什么问题。我花了很长时间思考这个，我看不出问题。如果有人可以看一看并指出错误，或者向我保证实际上没有问题，我真的很感激。

我认为语法在线程包函数（minithreads和semaphores）方面应该是非常标准的，但是如果有什么问题让我知道。

#include <stdio.h>
#include <stdlib.h>
#include "minithread.h"
#include "synch.h"
#define BUFFER_SIZE 16
#define MAXCOUNT 100

int buffer[BUFFER_SIZE];
int size, head, tail;
int count = 1;
int out = 0;
int toadd = 0;
int toremove = 0;

semaphore_t empty;
semaphore_t full;
semaphore_t count_lock; // Semaphore to keep a lock on the 
                        // global variables for maintaining the counts


/* Method to handle the working of a student 
    * The ID of a student is the corresponding minithread_id */
int student(int total_burgers) {
    int n, i;
    semaphore_P(count_lock);
    while ((out+toremove) < arg) {
        n = genintrand(BUFFER_SIZE);
        n = (n <= total_burgers - (out + toremove)) ? n : total_burgers - (out + toremove);
        printf("Student %d wants to get %d burgers ...\n", minithread_id(), n);
        toremove += n;
        semaphore_V(count_lock);
        for (i=0; i<n; i++) {
            semaphore_P(empty);
            out = buffer[tail];
            printf("Student %d is taking burger %d.\n", minithread_id(), out);
            tail = (tail + 1) % BUFFER_SIZE;
            size--;
            toremove--;
            semaphore_V(full);
        }
        semaphore_P(count_lock);
    }
    semaphore_V(count_lock);
    printf("Student %d is done.\n", minithread_id());
    return 0;
}

/* Method to handle the working of a cook 
    * The ID of a cook is the corresponding minithread_id */
int cook(int total_burgers) {
    int n, i;
    printf("Creating Cook %d\n",minithread_id());
    semaphore_P(count_lock);
    while ((count+toadd) <= arg) {
        n = genintrand(BUFFER_SIZE);
        n = (n <= total_burgers - (count + toadd) + 1) ? n : total_burgers - (count + toadd) + 1;
        printf("Cook %d wants to put %d burgers into the burger stack ...\n", minithread_id(),n);
        toadd += n;
        semaphore_V(count_lock);
        for (i=0; i<n; i++) {
            semaphore_P(full);
            printf("Cook %d is putting burger %d into the burger stack.\n", minithread_id(), count);
            buffer[head] = count++;
            head = (head + 1) % BUFFER_SIZE;
            size++;
            toadd--;
            semaphore_V(empty);
        }
        semaphore_P(count_lock);
    }
    semaphore_V(count_lock);
    printf("Cook %d is done.\n", minithread_id());
    return 0;
}

/* Method to create our multiple producers and consumers
    * and start their respective threads by fork */
void starter(int* c){
    int i;
    for (i=0;i<c[2];i++){
        minithread_fork(cook, c[0]);
    }
    for (i=0;i<c[1];i++){
        minithread_fork(student, c[0]);
    }
}


/* The arguments are passed as command line parameters
    * argv[1] is the no of students
    * argv[2] is the no of cooks */
void main(int argc, char *argv[]) {
    int pass_args[3];
    pass_args[0] = MAXCOUNT;
    pass_args[1] = atoi(argv[1]);
    pass_args[2] = atoi(argv[2]);

    size = head = tail = 0;
    empty = semaphore_create();
    semaphore_initialize(empty, 0);
    full = semaphore_create();
    semaphore_initialize(full, BUFFER_SIZE);
    count_lock = semaphore_create();
    semaphore_initialize(count_lock, 1);

    minithread_system_initialize(starter, pass_args);
}

Answer 1

您的信号量无法保护最内层循环中的缓冲区，磁头等。一个线程获得sempahore“empty”，另一个获取信号量“full”而没有其他信号量被保持。这似乎可以保证最终的腐败。

“信号量的使用是巧妙的错误”

1 个答案: