Posix线程与互斥锁同步

时间:2015-01-09 21:11:17

标签: linux multithreading

我越是尝试学习这个东西,我就越感到困惑,所以我愿意接受任何建议和帮助。谢谢。程序是关于窄桥(互斥)和运行汽车(进程)的,但是一次只能有一个进程交叉。在交叉之后,桥接线程可以将自己添加到队列或转到它睡觉的城市,然后添加到队列。进程(汽车)需要运行直到程序终止。如有必要,我可以将代码翻译成英文。编译后运行如下: ./ program -n -debug n - 线程数,调试 - 打印队列,仅可选。我认为线程不能同步工作,例如我为8个线程运行程序并且队列中有34号线程。不知道为什么,它发生在我“修复”代码之后。

/*
There's a road(bridge) from city A to B. Only one car can use it to move at a time. 
Cars should run( change cities ) all the time, no end of program.
Access to the bridge should be synchronized with mutexes. Every car have its numer from 1 to N where N is given as first parameter.
Program should printing something like this when one of printing variable is changing:
A - 5 10 >> >[>> 4 >> ] << <4 6 - B
@up That means that in city A is 5 cars, in city A queue is 10 cars ( wanting to change city ). Thread with numer 4 is changing city from A to B. In city B queue is 4 cars and in city B is 6 cars.
If second parameter -debug is given, then program should printf queue status.   
*/

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>
#include <errno.h>
#include <string.h>

#define true 1
#define false 0

pthread_mutex_t countOfQueueA;
pthread_mutex_t countOfQueueB; 
pthread_mutex_t bridge; 
pthread_mutex_t queueA; 
pthread_mutex_t queueB;
pthread_mutex_t cityA; 
pthread_mutex_t cityB; 

int inQueueA = 0; // Number of cars in queue of city A
int inQueueB = 0; // Number of cars in queue of city B
int inCityA = 0; // Number of cars in city A
int inCityB = 0; // Number of cars in city B

int createdThreads = 0; // Number of created threads by pthread_create
int numberOfCars = 0; // 
int debugMode = 0; // 0 - no, 1 - debug
int *queueInA; // Pointer to queue in city A
int *queueInB; // Pointer to queue in city B



void printQueue(char city)
{
    int i;
    if (city == 'a')
    {
    printf("\nQueue A status: ");
    for (i = 0; i<numberOfCars; i++)
    {
        printf("%d ", queueInA[i]);
    }
    printf("\n");
    }
    else if (city == 'b')
    {
    printf("\nQueue B status: ");
    for (i = 0; i<numberOfCars; i++)
    {
        printf("%d ", queueInB[i]);
    }
    printf("\n");
    }
}

void addToQueue(char city, int threadNumber) // Adding at the end of the queue in selected city
{
    if (city == 'a')
    {
    pthread_mutex_lock(&queueA);
    pthread_mutex_lock(&countOfQueueA);
    int i = 0;
    while (queueInA[i] != 0) //Looking for first free place = 0 to add car
    {
        i++;
    }
    queueInA[i] = threadNumber;
    inQueueA++;
    printf("\nA-%d %d>>> [ BLANK  ] <<<%d %d-B", inCityA, inQueueA, inQueueB, inCityB);
    if (debugMode == 1)
    {
        printQueue(city);
    }
    pthread_mutex_unlock(&queueA);
    pthread_mutex_unlock(&countOfQueueA);
    }
    else if (city == 'b')
    {
    pthread_mutex_lock(&queueB);
    pthread_mutex_lock(&countOfQueueB);
    int i = 0;
    while (queueInB[i] != 0) 
    {
        i++;
    }
    queueInB[i] = threadNumber;
    inQueueB++;
    printf("\nA-%d %d>>> [ BLANK  ] <<<%d %d-B", inCityA, inQueueA, inQueueB, inCityB);
    if (debugMode == 1)
    {
        printQueue(city);
    }
    pthread_mutex_unlock(&queueB);
    pthread_mutex_unlock(&countOfQueueB);
    }
}

void changeCity2(int threadNumber, char city)
{
    if (city == 'a')
    {
    while (queueInA[0] != threadNumber);// Oczekiwanie dopoki samochod nie jest 1szy w kolejce

    pthread_mutex_lock(&bridge);
    removeFromQueue(city, threadNumber);

    printf("\nA-%d %d>>> [>> %d  >>] <<<%d %d-B", inCityA, inQueueA, threadNumber, inQueueB, inCityB);
    if (debugMode == 1)
    {
        pthread_mutex_lock(&queueA);
        printQueue(city);
        pthread_mutex_unlock(&queueA);
    }
    city = 'b';

    pthread_mutex_unlock(&bridge);
    sleep(2); // Sleeping for simulating "working" time

    int randomNumber = rand() % 4;

    if (randomNumber % 2 == 0)
    {
        addToQueue(city, threadNumber);
        changeCity2(threadNumber, city);
    }
    else
    {
        runIntoCity(threadNumber, city);
    }
    }
    else if (city == 'b')
    {
    while (queueInB[0] != threadNumber); // Oczekiwanie dopoki samochod nie jest 1szy w kolejce      

    pthread_mutex_lock(&bridge);


    removeFromQueue(city, threadNumber);


    printf("\nA-%d %d>>> [<< %d  <<] <<<%d %d-B", inCityA, inQueueA, threadNumber, inQueueB, inCityB);
    if (debugMode == 1)
    {
        pthread_mutex_lock(&queueB);
        printQueue(city);
        pthread_mutex_unlock(&queueB);
    }
    city = 'a';


    pthread_mutex_unlock(&bridge);
    sleep(2);

    int randomNumber = rand() % 4;

    if (randomNumber % 2 == 0)
    {
        addToQueue(city, threadNumber);
        changeCity2(threadNumber, city);
    }
    else
    {
        runIntoCity(threadNumber, city);
    }
    }
}

void runIntoCity(int threadNumber, char city)
{

    if (city == 'a')
    {
    pthread_mutex_lock(&cityA);
    inCityA++;
    printf("\nA-%d %d>>> [ BLANK  ] <<<%d %d-B", inCityA, inQueueA, inQueueB, inCityB);
    if (debugMode == 1)
    {
        pthread_mutex_lock(&queueA);
        printQueue(city);
        pthread_mutex_unlock(&queueA);
    }
    pthread_mutex_unlock(&cityA);

    sleep(3);

    pthread_mutex_lock(&cityA);
    inCityA--;
    printf("\nA-%d %d>>> [ BLANK  ] <<<%d %d-B", inCityA, inQueueA, inQueueB, inCityB);

    if (debugMode == 1)
    {
        pthread_mutex_lock(&queueA);
        printQueue(city);
        pthread_mutex_unlock(&queueA);
    }
    pthread_mutex_unlock(&cityA);
    }
    else
    {
    pthread_mutex_lock(&cityB);
    inCityB++;
    printf("\nA-%d %d>>> [ BLANK  ] <<<%d %d-B", inCityA, inQueueA, inQueueB, inCityB);

    if (debugMode == 1)
    {
        pthread_mutex_lock(&queueB);
        printQueue(city);
        pthread_mutex_unlock(&queueB);
    }
    pthread_mutex_unlock(&cityB);

    sleep(3);

    pthread_mutex_lock(&cityB);
    inCityB--;
    printf("\nA-%d %d>>> [ BLANK  ] <<<%d %d-B", inCityA, inQueueA, inQueueB, inCityB);
    if (debugMode == 1)
    {
        pthread_mutex_lock(&queueB);
        printQueue(city);
        pthread_mutex_unlock(&queueB);
    }
    pthread_mutex_unlock(&cityB);
    }


    addToQueue(city, threadNumber);
    changeCity2(threadNumber, city);
}

void removeFromQueue(char city, int threadNumber) // Removing car from queue if its 1st in queue
{
    if (city == 'a') // Car being removed from queue of city A
    {
    pthread_mutex_lock(&queueA);
    pthread_mutex_lock(&countOfQueueA);
    if (queueInA[0] == threadNumber)
    {

        inQueueA--;

        int i = 1; 
        while (queueInA[i] != 0)
        {
            queueInA[i - 1] = queueInA[i];
            i++;
        }
        queueInA[i - 1] = 0;
        printf("\nA-%d %d>>> [ BLANK  ] <<<%d %d-B", inCityA, inQueueA, inQueueB, inCityB);


        if (debugMode == 1)
        {

            printQueue(city);
        }
    }
    else printf("Car is not first in queue. Error!");
    pthread_mutex_unlock(&queueA);
    pthread_mutex_unlock(&countOfQueueA);
    }
    else if (city == 'b') 
    {
    pthread_mutex_lock(&queueB);
    pthread_mutex_lock(&countOfQueueB);
    if (queueInB[0] == threadNumber)
    {

        inQueueB--;

        int i = 1;
        while (queueInB[i] != 0)
        {
            queueInB[i - 1] = queueInB[i];
            i++;
        }
        queueInB[i - 1] = 0;

        printf("\nA-%d %d>>> [ BLANK  ] <<<%d %d-B", inCityA, inQueueA, inQueueB, inCityB);

        if (debugMode == 1)
        {
            printQueue(city);
        }
    }
    else printf("Samochod nie jest pierwszy w kolejce. BLAD W KODZIE!");
    pthread_mutex_unlock(&queueB);
    pthread_mutex_unlock(&countOfQueueB);
    }
}



void changeCity(int threadNumber, char city)
{
    if (city == 'a')
    {
    while (queueInA[0] != threadNumber); // Waiting until car is ready to change city - it's first in queue

    pthread_mutex_lock(&bridge);
    removeFromQueue(city, threadNumber);



    printf("\nA-%d %d>>> [>> %d  >>] <<<%d %d-B", inCityA, inQueueA, threadNumber, inQueueB, inCityB);
    if (debugMode == 1)
    {
        pthread_mutex_lock(&queueA);
        printQueue(city);
        pthread_mutex_unlock(&queueA);
    }
    city = 'b';


    pthread_mutex_unlock(&bridge);
    sleep(2);

    int randomNumber = rand() % 4;

    if (randomNumber % 2 == 0)
    {
        addToQueue(city, threadNumber);
        changeCity2(threadNumber, city);
    }
    else
    {
        runIntoCity(threadNumber, city);
    }
    }
    else if (city == 'b')
    {
    while (queueInB[0] != threadNumber); // Waiting until car is ready to change city - it's first in queue  

    pthread_mutex_lock(&bridge);


    removeFromQueue(city, threadNumber);


    printf("\nA-%d %d>>> [<< %d  <<] <<<%d %d-B", inCityA, inQueueA, threadNumber, inQueueB, inCityB);
    if (debugMode == 1)
    {
        pthread_mutex_lock(&queueB);

        printQueue(city);
        pthread_mutex_unlock(&queueB);
    }
    city = 'a';


    pthread_mutex_unlock(&bridge);
    sleep(2);

    int randomNumber = rand() % 4;

    if (randomNumber % 2 == 0)
    {
        addToQueue(city, threadNumber);
        changeCity2(threadNumber, city);
    }
    else
    {
        runIntoCity(threadNumber, city);
    }
    }
}

char cityDraw() // Being used at start of thread to attach threads to cities
{
    int randomNumber = rand() % 100;
    int randomNumber2 = rand() % randomNumber;

    if (randomNumber2 % 2 == 0)
    {
    return 'a';
    }
    else
    {
    return 'b';
    }
}

void *threadInitiate(int threadNumber)
{
    char city = cityDraw();
    addToQueue(city, threadNumber);
    while (inQueueA + inQueueB < numberOfCars); // Waiting for all threads to get run by pthread_create
    changeCity(threadNumber, city);
}

void createThreads()
{
    pthread_t car[numberOfCars];
    int i;
    for (i = 0; i < numberOfCars; i++)
    {
    int wynik = pthread_create(&car[i], NULL, &threadInitiate, (void *)i + 1); //!!!!!!!!!!!!!
    if (wynik != 0) printf("Pthread_create failed\n");
    else createdThreads++;
    }

    for (i = 0; i < numberOfCars; i++)
    {
    pthread_join(car[i], NULL);
    }
}

void initiateQueues() // Making every elem. of queues became 0. Just to be sure. Thread numbers are starting from number 1.
{
    int i;
    for (i = 0; i<numberOfCars; i++)
    {
    queueInA[i] = 0;
    queueInB[i] = 0;
    }
}

int checkNumberOfCars(char *arg) // Parsing and converting to int,  numer of cars from parameter
{
    int argSize = 1;
    while (arg[argSize] != '\0')
    {
    argSize++;
    }

    char temp[argSize];
    int indArg = 1;
    int indTemp = 0;
    for (indArg = 1; indArg<argSize; indArg++)
    {
    temp[indTemp] = arg[indArg];
    indTemp++;
    }
    temp[indTemp] = '\0';

    int ls = atoi(temp);
    return ls;
}

int debugCheck(int argc, char **argv) // Checking if -debug parameter is given
{
    if (argc>2)
    {
    if (strcmp(argv[2], "-debug") == 0)
        return true;
    else
        return false;
    }
}

int main(int argc, char **argv)
{
    numberOfCars = checkNumberOfCars(argv[1]);
    printf("\nNumber of cars from param = %d", numberOfCars);
    debugMode = debugCheck(argc, argv);
    if (debugMode == 1) printf("\nDebugMode is ON - writing queues status on every change");
    int queueArrayA[numberOfCars];
    int queueArrayB[numberOfCars];
    queueInA = queueArrayA;
    queueInB = queueArrayB;
    initiateQueues();

    pthread_mutex_init(&bridge, NULL);
    pthread_mutex_init(&queueA, NULL);
    pthread_mutex_init(&queueB, NULL);
    pthread_mutex_init(&cityA, NULL);
    pthread_mutex_init(&cityB, NULL);
    pthread_mutex_init(&countOfQueueA, NULL);
    pthread_mutex_init(&countOfQueueB, NULL);

    createThreads();

    return 0;
}

1 个答案:

答案 0 :(得分:0)

这看起来像是并发编程课程的作业。这段代码存在很多问题,我不认为值得尝试找到错误,因为你不会得到任何积分,无论如何都必须重写。以下是大致按重要性排列的问题。

  1. 队列的目的是什么?汽车线程应该都试图锁定互斥锁。程序中不需要队列 - 操作系统在后台管理等待互斥锁的线程队列。汽车线程应该做这样的事情:

    pthread_mutex_lock(&bridge);
    // remove car from city of origin
    // no need for mutex, because only one car can be on the bridge at once,
    // so only the car on the bridge will modify the car count fields in cities
    city[i].cars--;
    sleep(/* some time */);
    // add car to destination city
    city[1-i].cars++;
    pthread_mutex_unlock(&bridge);
    sleep(/* some time */);
    
  2. 汽车线程将永远运行,但他们将继续以递归方式调用越来越多的函数。这最终将炸毁堆栈。将汽车线程函数写为显式无限循环。没有物理计算机可以让你拥有无限深度的调用图。
  3. 大量的重复代码。一切都写了两次,一次是针对城市&#39; a&#39;第二次来到城市&#39; b&#39;。利用数组并将数字0分配给一个城市,将1分配给另一个城市,以便您可以执行city[1-i].car_count++之类的操作。
  4. 管理队列数据结构的代码(例如,查找空白空间,从队列中删除汽车等)与其他不相关的代码混合在一起。如果你真的需要那些我怀疑的队列,写一个struct car_queue和添加和删除汽车的操作,而不是直接在函数changeCity2addToQueue中编写队列操作等等
  5. 不必要的互斥锁。例如,countofQueueAqueueA应合并为一个互斥锁。事实上,可能只有三个互斥体:城市A,城市B和桥梁。它甚至可以只用一个互斥桥来完成。
  6. 为什么你在rand() % 4之后和% 2之后正在做结果?它没有任何意义,马上做rand() % 2
  7. cityDraw中的双重随机化是完全没有意义的,会给你一个稍微有偏见的结果。例如,如果你在第一行得到0,你将在第二行获得0,1或2,所以将有2/3的机会将汽车转到城市A.如果你在第一行得到0无论你在第二场比赛中得到什么,赛车都将前往A市。只需rand() % 2
  8. 永远不要做像#define true 1这样的事情 - 这只是引入了与C ++的无偿不兼容性。相反,请使用#include <stdlib.h>并撰写TRUEFALSE